Simple Harmonic Motion notes class 11

Introduction

A motion which repeats itself after a definite interval of time is called periodic motion. The time interval after which the motion repeats is called the time period. Circular motion is one example of periodic motion which we come across everyday like the hands of a clock undergo circular motion with different time periods. Similarly, if a body moves to and fro on the same path, it is said to perform oscillations. Simple harmonic motion is a type of oscillatory motion in which the displacement x of the particle from the origin is given by

x = Asin(ωt +ф)

where A, ω and ф are constants. This kind of motion where displacement is a sinusoidal function of time is called simple harmonic motion. The quantities A, ω and ф which determine the shape of the graph (displacement curve) are:

Amplitude

The quantity A is called the amplitude of the motion. As sin(ωt +ф) can take values between -1 and +1, the displacement x can take values between –A and +A. This gives the physical significance of the constant A. It is the maximum displacement of the particle from the center of oscillation i.e. the amplitude of oscillation.

Phase Angle (ф)

The time varying quantity (ωt +ф) is called the phase of the motion and the quantity ф is called the phase angle or phase constant. It depends on the choice of the instant t = 0. To describe the motion quantitatively, a particular instant should be called zero and measurement of time should be made from this instant. If we choose to call the instant when the particle is passing through its mean position as the time t = 0, then ф = 0 as at that time x has to be zero. But on the other hand, if for the convenience of the problem, t = 0 has to be considered at a point when the particle at an extreme position, then ф would be ±п/2 depending on the direction of velocity of the particle at that position.

Angular frequency (ω)

To understand angular frequency, we first try to understand frequency of simple harmonic motion. We know that the motion repeats after a certain time called the time period (T) of the motion i.e x should have the same value at time t and t+T.

Thus, sin(ωt + ф) = sin(ω(t+T) + ф)

Also, the velocity should have the same value at t+T, i.e.

ωcos(ωt + ф) = ωcos(ω(t+T) + ф)

As T is the smallest time for repetition both the above equations are true for ωT = 2п, n = 1,2,3,……

or ω = 2п/T

As (1/T) is the frequency of oscillation i.e. the number of oscillations per unit time, (2п/T) is the angular frequency i.e. the number of radians covered per unit time (where 1 complete oscillation corresponds to 2п radians).

Velocity

The expression for velocity can be derived by differentiating the displacement w.r.t. time i.e.

v(t) = dx(t)/dt

= d[Asin(ωt + ф)]/dt

= Aωcos(ωt + ф)

From the expression it can be seen that the velocity varies between the extreme values –Aω and +Aω. Also as we know that sin(90 + θ) = cosθ, we can conclude that the velocity curve is shifted to the left by 90o w.r.t. the displacement curve. When the magnitude of displacement is least, that of the velocity is maximum and vice versa.

Acceleration

Knowing the velocity we can arrive at acceleration for simple harmonic motion by differentiating the velocity, i.e.

a(t) = dv(t)/dt

= d[Aωcos(ωt + ф)]/dt

= -Aω2sin(ωt + ф)

Thus the acceleration of the particle varies between the limits -Aω2 and +Aω2. Also the acceleration curve is shifted to the left by 90o w.r.t. the velocity curve.

Combining the equations for acceleration and displacement, we get

a(t) = -ω2x(t)

which implies that the acceleration is proportional to the displacement in simple harmonic motion and the two related by the square of the angular frequency.

Force Law for SHM

From Newton’s second law we know that F = ma and that for SHM, a = -Aω2sin(ωt + ф). Thus, for simple harmonic motion, F = -mAω2sin(ωt + ф) = -mω2x(t)

This force law is familiar. It is Hooke’s law

F = -kx where k = mω2

For a spring, spring constant being k = mω2

Thus the spring-block system forms a simple harmonic oscillator with angular frequency, ω = √(k/m) and time period, T = 2п/ω = 2п√(m/k).

Energy of SHM

Simple Harmonic motion is defined by the equation F = -kx. The work done by the force F during a displacement from x to x + dx is

dW = Fdx

= -kxdx

The work done in a displacement from x = 0 to x is

W = 0x-kxdx = -kx2/2

As the change in potential energy corresponding to a force is negative of the work done by the force, U(x) – U(0) = -W = ½ kx2

Let us choose the origin to be zero of potential energy, then U(0) = 0 and U(x) = ½ kx2

The kinetic energy at a time t is K = ½ mv2

The total mechanical energy at time t is E = U + K

= ½ kx2 + ½ mv2 = ½ kA2sin2(ωt + ф) + ½ mA2ω2cos2(ωt + ф)

Putting k = mω2, we get

E = ½ mω2[A2sin2(ωt + ф) + cos2(ωt + ф)] = ½ mω2A2

Thus the total mechanical energy remains constant and is independent of time.

Simple Pendulum

A simple pendulum consists of a heavy particle suspended from a fixed support through a light, inextensible string. This system can stay in equilibrium if the string is vertical. This is called the mean position or the equilibrium position. If the particle is pulled aside and released, it oscillates in a circular arc with the center at the point of suspension ‘O’.

At angle θ, the forces acting on the particle are:

(1) the weight mg downwards

(2) The tension T in the string

As the particle makes pure rotation about a horizontal axis through O (say OA), let us find the torques of the forces acting on it.

The torque of T is zero as it intersects OA at point O.

The torque of mg about OA is

τ = mg(l sinθ)

As this torque tries to bring the particle back towards θ = 0, we can write

τ = -mgl sinθ

According to this equation, the resultant torque is not proportional to the angular displacement, hence the motion is not angular simple harmonic.

However, if the angular displacement is small, sinθ is approximately equal to θ (expressed in radians) and the equation may be written as

τ = -mgl θ

Thus, if the amplitude of oscillation is small, the motion of the particle is approximatelt angular simple harmonic.

The moment of inertia of the particle about the rotation axis OA is ml 2

Hence, angular acceleration, α = τ/I = -mgl θ/ml 2 = -(g/l )θ

Also, in SHM, α = -ω2θ

Therefore, ω = √(g/l )

and the time period is T = 2п/ω = 2п√(l /g).

Physical Pendulum

Any rigid body suspended from a fixed support constitutes a physical pendulum. The motion of such a system can be followed very nearly in the same way as that of a simple pendulum. The distance of the center of mass of the body from the fixed suspension point acts as the effective length of the pendulum and the total mass being the mass of the particle situated at the center of the body. This system can now be treated in exactly the same way as a simple pendulum.

The restoring torque acting on the body when rotated through an angle θ is

τ = -mgl sinθ

Let the moment of inertia of the body about the horizontal rotation axis through point of suspension be ‘I’. Thus, angular acceleration of the body is

α = τ/I = -(mgl /I)sinθ

For small displacements, sinθ ≈ θ, therefore,

α = -(mgl /I)θ

Hence the body undergoes simple harmonic oscillations, with angular frequency, ω = √(mgl /I) and time period, T = 2п/ω = 2п√(I/mgl )

Torsional Pendulum

In a torsional pendulum, an extended body is suspended by a light wire. The body is rotated about the wire as the axis of rotation. The upper end of the wire remains fixed with the support and the lower end of the wire is rotated through an angle with the body, thus a twist θ is produced in the wire.

The twisted wire exerts a restoring torque on the body to bring it back to its original position i.e. θ = 0. This torque has a magnitude proportional to the angle of twist, θ, i.e. τ = -kθ, where k is called the torsional constant of the wire.

If the moment of inertia of the body about the vertical axis is ‘I’,

the angular acceleration, α = τ/I = -(k/I)θ

Thus the motion of the body is simple harmonic with angular frequency, ω = √(k/I) and time period, T = 2п/ω = 2п√(I/k)

Damped Oscillations

When the motion of an oscillator is reduced by an external force, the oscillator and its motion are said to be damped. The damping force is a function of speed and is directed opposite to the velocity. Energy is lost due to the negative work done by the damping force and the system comes to a halt in due course.

The damping may be a complicated function of speed but in several cases of practical interest the damping force is proportional to the speed and can be written as

F = -bv

The equation of motion becomes

F = -kx –bv

or mdv/dt = -kx –bv

or md2x/dt2 = -kx – bdx/dt

or md2x/dt2 + bdx/dt + kx = 0

The solution of this differential equation is x(t) = Ae-bt/2msin(ω′t + δ)

where ω′ = √((k/m) – (b/2m)2) = √(ω02 – (b/2m)2)

For small b, the angular frequency ω′ ≈ √(k/m) = ω0. Thus, the system oscillates with almost the natural angular frequency and with amplitude decreasing with time according to

A = A0e-bt/2m

The amplitude decreases and finally becomes zero.

The energy of a damped oscillator (if the damping is small) can be approximated as

E(t) = ½ mω2A02e-bt/m = ½ kA02e-bt/m

Problems

Q:1 A loudspeaker produces a musical sound by means of the oscillation of a diaphragm whose amplitude is limited to 1.00 μm. (a) At what frequency is the magnitude a of the diaphragm’s acceleration equal to g? (b) For greater frequencies, is ‘a’ greater than or less than g? Take g = 10 ms-2.

Solution: The magnitude of maximum acceleration is given by ω2xm where ω is the angular frequency and xm is the amplitude of oscillation.

(a) The angular frequency for which maximum acceleration is g, is

ω = √(g/xm)

The corresponding frequency = ω/2п = (1/2п)√(10/10-6) = 503 Hz

(b) For greater frequencies, maximum acceleration is greater than ‘g’ but minimum acceleration is always zero at the mean position. Thus ‘a’ exceeds ‘g’ for some part of the motion and is less than ‘g’ for the rest.

Q:2 What is the phase constant for the harmonic oscillator with the velocity function v(t) given in figure if the position function x(t) has the form x = Acos(ωt +ф)? The vertical axis scale is set by vs = 4 cm/s.

Solution: We know that vm = ωA where A is the amplitude of oscillation.

From the graph we can see that vm = ωA = 5 cm/s

Also, at t =0, v = 4 cm/s

Now, given that x = Acos(ωt +ф)

We know that v = dx/dt

Therefore, v = -Aωsin(ωt +ф)

At t = 0,

4 = -5sinф

or ф = sin-1(-4/5) = -53o , 233o

As can be seen from the graph, acceleration at t=0 is negative (because slope of the graph is negative)

a = dv/dt = -Aω2cos(ωt +ф)

Thus, ф = -53o because the other value makes ‘a’ positive at t = 0.

Q:3 The equation of motion of a particle started at t=0 is given by x = 10sin(10t + п/3), where x is in centimeter and t in second. When does the particle (a) first come to rest (b) first have zero acceleration (c) first have maximum speed?

Solution: The particle first comes to rest at the maximum displacement from the mean position i.e. the amplitude of the oscillation.

Thus 10 = 10sin(10t + п/3)

or sin(10t + п/3) = 1

10t = п/2 – п/3 = п/6

which gives t = п/60 s

(b) We know that, a = ω2x = ω210sin(10t + п/3)

For a = 0, ω210sin(10t + п/3) = 0

or 10t + п/3 = п

or t = п/15 s

v is maximum when cos(10t + п/3) = 1 or -1

cos(10t + п/3) becomes -1 before it becomes 1, so

(10t + п/3) = п

or 10t = 2п/3

or t = п/15 s

Q:4 Figure shows block 1 of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 8 m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant 1200 N/m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of 0.2 s, and block 1 slides off the opposite end of the elevated surface, landing a distance d from the base of that surface after falling height h = 5 m. What is the value of d? Use g = 10 ms-2.

Solution: We know that time period of a simple harmonic motion, T = 2п√(m/k)

Therefore, 2п√(m2/k) = 0.2

which gives m2 = 0.01 X 1200/п = 1.2 kg

Now, we can find the rebound speed of block 1 after collision,

v1f = (0.2-1.2)/(0.2+1.2) X 8 = -5.7 m/s i.e. 5.7 m/s towards the left.

As the surface is frictionless, this is the initial velocity of the projectile.

Time of flight = √(2h/g) = 1 s

Therefore, d = 5.7 X 1 = 5.7 m

Q:5 Find the period of oscillation of mass ‘m’ in the given figures. What is the equivalent spring constant of the pair of springs in each case?

Solution: (a) If the block is displaced from its equilibrium position to the right by a distance ‘x’, the restoring force acting on it by the two springs is F = -k1x – k2x = -(k1 + k2)x

or F = -kx where k = (k1 + k2) = equivalent spring constant for the given system.

Now, acceleration, a = -(k1 + k2)x/m

Thus, ω2 = (k1 + k2)/m

Hence, time period, T = 2п√(m/(k1 + k2))

(b) Again, if the block is displaced by a distance ‘x’ towards the right, the total force exerted by the two springs is F = -k1x – k2x

Thus the equivalent spring constant for the system is k = (k1 + k2).

Acceleration of the block, a = -(k1 + k2)x/m

Thus, ω2 = (k1 + k2)/m

Hence, time period, T = 2п√(m/(k1 + k2))

Let the spring with spring constant k1 be extended by distance x1 and the spring with spring constant k2 be extended by distance x2 with x1 + x2 = x …(1)

As the net force between the springs must be zero, the force exerted by two springs must be equal, so that they do not accelerate w.r.t. each other.

Thus, we have,

k1x1 = k2x2

or x1 = (k2/k1)x2 …(2)

Putting (2) in (1);

(k2/k1)x2 + x2 = x

or x2 = k1x/(k1 + k2)

Similarly, x1 = k2x/(k1 + k2)

Now, the force exerted on the block = the force exerted by any of the springs.

It is important to understand here that the two forces do not add up because if it were so, the springs would accelerate w.r.t each other. Suppose the force exerted by the spring connected to the block is (F1(for spring 1) + F2(for spring 2)), the block also exerts the same force on the end of the spring connected to it whereas the other end of the spring is only pulled by a force F1(from spring 1), this would result in the spring 2 having a resultant force acting on it which is not possible.

Thus, the force on the block is F = k1x1 = k2x2

or F = k1k2x/(k1 + k2)

Acceleration of the block, a = k1k2x/m(k1 + k2)

Thus, time period of oscillations, T = 2п√[m(k1 + k2)/k1k2] and equivalent spring constant,

k = k1k2/(k1 +k2).

Q:6 Consider a simple harmonic motion of time period T. Calculate the time taken for the displacement to change value from half the amplitude to the amplitude (b) velocity to change value from half its maximum to maximum.

Solution: (a) We know that x = Asinωt = Asin(2п/T)t where A is the amplitude of the oscillation.

Time to reach A/2 from mean position:

A/2 = Asin(2п/T)t1

or п/6 = (2п/T)t1

or t1 = T/12

Time taken to reach the extreme position:

A = Asin(2п/T)t2

or п/2 = (2п/T)t2

or t2 = T/4

Therefore, time taken for the displacement to change value from half the amplitude to the amplitude = t2 – t1 = T/4 – T/12 = T/6

(b) We know that v = Aωcosωt = A(2п/T)cos(2п/T)t

Velocity is maximum at the mean position, at t=0 and the maximum velocity = A(2п/T)

Time taken for the velocity to become half its maximum:

Aп/T = A(2п/T)cos(2п/T)t

½ = cos(2п/T)t

or п/3 = 2пt/T

or t = T/6

Therefore, time taken for the velocity to change value from half its maximum to maximum = 0-T/6= T/6

Q:7 A body of mass 2 kg suspended through a vertical spring executes simple harmonic motion of period 6 s. If the oscillations are stopped and the body hangs in equilibrium, find the potential energy stored in the spring. Use g = 10 ms-2.

Soltuion: Time period = 2п√(m/k)

6 = 2п√(2/k)

3/п2 = 2/k

k = 6.58 N/m

Now, when the oscillations stop, the mass is in equilibrium.

Therefore, mg = kx

20 = 6.58x

or x = 3.04 m

Thus, potential energy stored in the spring = ½ kx2 = ½ X 6.58 X (3.04)2 = 30.4 J

Q:8 In figure, two blocks (m = 2 kg and M = 12 kg) and a spring of k = 200 N/m are arranged on a horizontal, frictionless surface. The coefficient of static friction between the two blocks is 0.4. What amplitude of simple harmonic motion of the spring–blocks system puts the smaller block on the verge of slipping over the larger block? Use g = 10 ms-2.

Solution: On the verge of slipping means that the magnitude of force of friction is maximum i.e equal to μsmg at the time of maximum acceleration of the system.

We know that magnitude of maximum acceleration = Aω2 where A is the amplitude of oscillation. In this problem, ω = √(k/(m+M))

Therefore, mAω2 = μsmg

or Ak/(m+M) = μsg

or A = μsg(m + M)/k = 0.4 X 10 x 14/200 = 0.28 m = 28 cm

Q:9 The springs shown in figure are all unstretched in the beginning when a man starts pulling the block. The man exerts a constant force F on the block. Find the amplitude and the frequency of motion of the block.

Solution: The equivalent spring constant for springs 2 and 3 is k2k3/(k2 + k3). This equivalent spring is now in series connection with the spring 1, so the equivalent spring constant for the three springs is k = k1 + k2k3/(k2 + k3) = (k1k2 + k1k3 + k2k3)/(k2 + k3).

Therefore, time period, T = 2п√[m(k2 + k3)/(k1k2 + k1k3 + k2k3)].

Frequency = 1/T = (1/2п)√[(k1k2 + k1k3 + k2k3)/m(k2 + k3)]

Amplitude = F/k = F(k2 + k3)/(k1k2 + k1k3 + k2k3).

Q:10 A particle is subjected to two simple harmonic motions, one along the x-axis and the other on a line making an angle of 45o with the x-axis. The two motions are given by

x= x0sinωt and s = s0sinωt

Find the amplitude of the resultant motion.

Solution: The resultant motion will itself be simple harmonic in nature as the frequency of the individual motions is same.

The resultant amplitude can be found by using the vector method of combining two simple harmonic motions.

As the angle between the two motions is 45o, we can write

The resultant motion, R = [x0sinωt + (s0sinωt)cos45]i + (s0sinωt)sin45j

Using Pythagoras theorem,

R = sinωt√[(x0 + (s0/√2))2 + (s0/√2)2] = sinωt√[x02 + s02/2 + √2x0s0 + s02/2]

= √[x02 + s02 + √2x0s0] sinωt

Therefore, resultant amplitude = √(x02 + s02 + √2x0s0)

Q:11 In a damped oscillator with m = 500 g, k = 100 N/m, and b = 75 g/s, what is the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of 20 cycles?

Solution: As the value of b (0.075 kg/s) is very small as compared to √km (7.07 kg/s), the period of oscillations is approximately equal to 2п√(m/k) = 0.44 s

After 20 oscillations, t = 20T = 8.8 s,

the ratio of amplitude to the initial amplitude = e-bt/2m = 0.51.

Q:12 A physical pendulum consists of two meter-long sticks joined together as shown in figure. What is the pendulum’s period of oscillation about a pin inserted through point A at the center of the horizontal stick?

Solution: Let the mass of each rod be ‘m’. Length of each rod = 1m

Moment of inertia of horizontal rod about its center = ml 2/12 = m/12

Moment of inertia of the vertical rod about point A = ml 2/12 + ml 2/4 = ml 2/3 = m/3

Moment of inertia of the system about the axis of rotation = m/12 + m/3 = 5m/12

The center of mass of the system is at l /4 from point A. Let the system be displaced by an angle θ about the axis of rotation, the total torque acting on the system = -2mg(l /4)sinθ

The angular acceleration becomes α = τ/I = -mg(l /2)sinθ/I

For small displacements sinθ ≈ θ, so that

α = -mg(l /2)θ/I = -ω2θ

where ω2 = mgl /2I

Thus, T = 2п/ω = 2п√(2I/mgl )

Putting I = 5m/12 and l = 1 m, we get

T = 2п√(2(5m/12)/mg) = 2п/√(5/6g)

Q:13 A block of mass ‘m’ is suspended from the ceiling of a stationary standing elevator through a spring of spring constant ‘k’. Suddenly the cable breaks and the elevator starts falling freely. Show that the block now executes a simple harmonic motion in the elevator. Find the amplitude.

Solution: When the elevator is stationary, the spring is stretched and the weight of the block is supported by the spring force. Let the extension of the spring be ‘x’, we have kx = mg. Thus, x = mg/k.

As the cable breaks, the elevator starts falling freely with acceleration ‘g’. Let us work in the frame of reference of the elevator. As our frame is accelerating with ‘g’ downwards, we have to use a pseudo force ‘mg’ acting upwards on the block. Now, this force balances the weight of the block. Therefore, now there is a net force kx acting on the block upwards when it is at a distance ‘x’ from the position of the unstretched spring. Hence, its motion in the elevator is simple harmonic with the mean position corresponding to the unstretched spring. Initially, the spring is stretched by x = mg/k where the velocity of the block (P) is zero. Thus, the amplitude of the resulting simple harmonic motion is mg/k.

Q:14 The pulley shown in figure has a moment of inertia ‘I’ about its axis and mass ‘m’. Find the time period of vertical oscillation of its center of mass. The spring has spring constant ‘k’ and the string does not slip over the pulley.

Solution: Let us find the equilibrium position of the pulley. For the rotational equilibrium of the pulley, the tensions in the two strings must be equal for the total torque on the pulley to be zero. Let this tension be T. The spring will be extended by y = T/k, as the tension in the spring will be same as tension in the string. Now, for translational equilibrium of the pulley,

2T = mg or 2ky = mg or y = mg/2k

Thus, the spring is extended by a distance mg/2k when the pulley is in equilibrium.

Now, suppose the center of the pulley goes down further by a distance x. The total increase in the string plus the spring is 2x (i.e. x on both sides). As the string has a constant length, the extension of the spring is 2x.

Thus the energy of the system is U = ½Iω2 + ½mv2 – mgx + ½k((mg/2k) + 2x)2

= ½((I/r2) + m)v2 + m2g2/8k + 2kx2

As the system is conservative, dU/dt = 0

which gives [(I/r2) + m]v dv/dt + 4kx dx/dt = 0

or [(I/r2) + m] dv/dt + 4kx = 0

or dv/dt = -4kx/[(I/r2) + m]

or a = -ω2x, where ω2 = 4k/[(I/r2) + m]

Thus, the center of the pullry executes a simple harmonic motion with a time period

T = 2п√(I/r2 + m)/4k

Q:15 Find the time period of the motion of the particle shown in figure. Neglect the effect of the bend at the bottom. Use g = 10 ms-2.

Solution: Let the time taken to travel AB and BC be t1 and t2 respectively. As one oscillation is complete when the particle reaches the starting point A, we see that time period, T = 2(t1 + t2).

Now, for part AB (taking positive y-axis downwards), we have

u = 0, a = gsin45, AB = 0.1/sin45

Using s = ut + ½at2, we have

0.1√2 = 0 + (5/√2)t12

or t1 = 0.2 s

Now, using v = u + at, we get

v = 0 + 10/√2 X 0.2 = √2 m/s

Now, for part BC (taking positive y-axis upwards), we have

u = √2 m/s, a = -gsin30, v = 0

Using v = u + at, we have

0 = √2 - 10/2 X t2

or t2 = √2/5 s

Hence, time period, T = 2(0.2 + √2/5) = 0.97 s

Q:16 A uniform rod of mass ‘m’ and length ‘l’ is suspended through a light wire of length ‘l’ and torsional constant ‘k’ as shown in figure. Find the time period if the system makes (a)small oscillations in the vertical plane about the suspension point and (b) angular oscillations in the horizontal plane about the center of the rod.

Solution: (a) The oscillations take place about the horizontal line through the point of suspension and perpendicular to the plane of the figure. This is the behavior of a physical pendulum.

Suppose the rod is displaced by an angle θ from its equilibrium position. The total torque acting on the rod about the point of suspension of string (i.e. axis of rotation) = mgl sinθ. This torque tries to bring the rod back to θ = 0, thus we can write, τ = -mgl sinθ. We know that angular acceleration = τ/I = -(mgl /I)sinθ. Applying the small angle approximation, we get

α = -(mgl /I) θ = -ω2θ

where ω2 = mgl I-1. Thus time period = 2п/ω = 2п√(I/mgl )

Now, the moment of inertia of the rod about the rotation axis is

I = ml 2/12 + ml 2 = 13ml 2/12

The time period = 2п√(I/mgl ) = 2п√(13ml 2/12mgl ) = 2п√(13l /12g).

(b) The angular oscillations take place about the suspension wire. This is the behavior of a torsional pendulum.

Suppose the rod is rotated through an angle θ, the torque produced is τ = -kθ. The angular acceleration thus is α = τ/I = -(k/I)θ = -ω2θ

where ω2 = k/I. Thus the time period of the motion is T = 2п/ω = 2п/√(I/k)

Now, the moment of inertia of the rod about the rotation axis is ml 2/12.

The time period is 2п√(I/k) = 2п√(ml 2/12k).

Q:17 A simple pendulum is suspended from the ceiling of a car accelerating uniformly on a horizontal road. If the acceleration is ‘a0’ and the length of the pendulum is ‘l ’, find the time period of small oscillations about the mean position.

Solution: Let us work in the frame of reference of the car. As the car is accelerating w.r.t the road, we have to apply a pseudo force ma0 on the bob of the pendulum.

Let the mean position of the pendulum be at a position when the string makes an angle θ with the vertical. At this position, the tension by the string, weight and the pseudo force will add up to zero.

Now, suppose at some instant during oscillation, the string is further deflected by an angle α so that the displacement of the bob is x.

Taking the components perpendicular to the string,

component of T = 0,

component of mg = mgsin(α+θ)

component of ma0 = -ma0cos(α+θ)

Thus, the resultant component F = m[gsin(α+θ) - a0cos(α+θ)]

Expanding the sine and the cosine and putting cosα ≈ 1, sinα ≈ α = x/l , we get

F = m[gsinθ - a0cosθ + (gcosθ + a0sinθ)x/l ] …(1)

At x = 0, the force F on the bob should be zero, as this is the mean position. Thus by (1),

0 = m[gsinθ - a0cosθ] …(2)

which gives tanθ = a0/g

Thus, sinθ = a0/√(a02 + g2) …(3)

and cosθ = g/√(a02 + g2) ...(4)

Putting (2), (3) and (4) in (1);

F = m√(a02 + g2)x/l

or F = mω2x where ω2 = √(a02 + g2)/l

This is an equation for simple harmonic motion with time period T = 2п/ω= 2п√l /(a02 + g2)1/4

Q:18 A damped harmonic oscillator consists of a block (m = 2 kg), a spring (k = 30 N/m), and a damping force (F = -bv). Initially, it oscillates with an amplitude of 25 cm; because of the

damping, the amplitude falls to three-fourths of this initial value at the completion of four oscillations. (a) What is the value of b? (b) How much energy has been “lost” during these four oscillations?

Solution: (a) We assume that b is small compared to √km and we take T = 2п/√(m/k) ≈ 1.62 s. It is given that at t = 4T, the amplitude falls to 3A/4, i.e.

e-bt/2m = 3/4

-2bT/m = ln(3/4)

or b = 0.18 kg/s.

(b) Energy lost during these four oscillations = ½ k(A2 – (3A/4)2) = 7kA2/32 = 0.410 J

Q:19 What is the frequency of a simple pendulum 2 m long (a) in a room, (b) in an elevator accelerating upward at a rate of 2 m/s2, and (c) in free fall? Use g = 10 ms-2.

Solution: (a) The frequency of a simple pendulum for small amplitude of oscillations is

f = (1/2п)√(g/l) = 0.36 Hz

(b) The acceleration of the elevator is 2 m/s2 upwards. Considering the motion of the bob of the pendulum in the non-inertial elevator frame, we have to add a pseudo force of magnitude ma = 2m acting downwards.

Therefore, the total force acting on the bob downwards = (mg + ma) = (g + 2)m

Therefore, ω = √((g+2)/l )

Thus frequency of oscillation, f = (1/2п)√((g+2)/l ) = 0.39 Hz

(c) Again, considering the motion in a non-inertial frame accelerating downwards with ‘g’, we have to use a pseudo force of magnitude mg acting on the bob upwards.

Therefore, the total force acting on the bob downwards = (mg - mg) = 0 which implies that the bob does not oscillate at all in free fall.

Q:20 A uniform disc of radius ‘r’ is to be suspended through a small hole made in the disc. Find the minimum possible time period of the disc for small oscillations. What should be the distance of the hole from the center of it to have minimum time period?

Solution: Let the distance of the hole from the center of mass of the disc be ‘x’. Let mass of disc be ‘m’.

Moment of inertia of the disc about the rotation axis = ½ mr2 + mx2 = m(r2/2 + x2)

We know that time period of a physical pendulum is T = 2п√(I/mgl ) = 2п√[m(r2/2 + x2)/mgx]

= 2п√[(r2 + 2x2)/2gx]

Now, for T to be minimum, dT/dx = 0

or d(T2)/dx = 0 as the square root of a number is minimum when the number itself is minimum.

Therefore, d[(4п2r2/2gx) + (8п2x2/2gx)]/dx = 0

or -4п2r2/2gx2 + 8п2/2g = 0

or -r2/x2 + 2 = 0

or x = r/√2

Putting in the expression for T, we get

T = 2п√[(r2 + 2(r/√2)2)/2gr/√2] = 2п√[(2√2r2)/2gr] = 2п√[(√2r)/g]

Hence the distance of the hole from the center of it to have minimum time period is r/√2 and the minimum time period is T = 2п√(√2r/g).

Q:21 A small block oscillates back and forth on a smooth concave surface of radius R. Find the time period of small oscillations.

Solution: Let the block be at an angle ‘θ’ with the vertical. The forces acting on the block are:

The force accelerating the block towards the mean position, F = mgsinθ.

As θ is very small, sinθ ≈ θ, therefore, F = mgθ or a = gθ

Let ‘x’ be the displacement of the block from its mean position,

Therefore, θ = x/R

Putting in the expression for ‘a’, we get

a = gx/R

As acceleration is proportional to ‘x’, the block undergoes simple harmonic motion.

Now, from the expression for ‘a’, we get ω2 = g/R

Therefore, time period, T = 2п/ω = 2п√(g/R)

Q:22 A rectangular block, with face lengths a = 35 cm and b = 45 cm, is to be suspended on a thin horizontal rod running through a narrow hole in the block. The block is then to be set swinging about the rod like a pendulum, through small angles so that it is in SHM. The given figure shows one possible position of the hole, at distance r from the block’s center, along a line connecting the center with a corner. (a) For what value of r does the pendulum have minimum time period? There is actually a line of points around the block’s center for which the period of swinging has the same minimum value. (b) What shape does that line make?

Solution: (a) For the physical pendulum, time period, T = 2п√(I/mgr)

The moment of inertia of the block about the given rotation axis = m(a2 + b2)/12 + mr2

Therefore, T = 2п√[(m(a2 + b2)/12 + mr2)/mgr] = 2п√[(a2 + b2)/12gr + (r/g)]

= (2п/√g)√[(a2 + b2)/12r + r]

For the time period to be minimum, dT/dr = 0

or d(T2)/dr = 0 as T will be minimum when its square is minimum.

Therefore, we have,

-(a2 + b2)/12r2 + 1 = 0

12r2 = a2 + b2

r = √[(a2 + b2)/12]

Putting in the values of a and b, we get

r = 0.165 m = 16.5 cm

(b) For the time period to have this minimum value, the distance of the hole should be equal to √[(a2 + b2)/12] irrespective of the direction. Therefore, the locus of points is a circle of radius

((a2 + b2)/12)1/2 from the centre.

Q:23 The given figure shows an overhead view of a long uniform rod of mass 1 kg which is free to rotate in a horizontal plane about a vertical axis through its center. A spring with force constant k = 500 N/m is connected horizontally between one end of the rod and a fixed wall. When the rod is in equilibrium, it is parallel to the wall. What is the period of the small oscillations that result when the rod is rotated slightly and released?

Solution: Let the length of the rod be ‘L’. We can see from the figure that if the torque exerted by the spring on the rod is proportional to the angle of rotation of the rod, then it will undergo simple harmonic motion.

Let l0 be the distance of the pivot point from the wall. This is also the equilibrium length of the spring. Let the rod rotate by an angle θ with the left end moving away from the wall. This end is now (L/2)sinθ further from the wall and has moved (L/2)(1-cosθ) towards the right.

The length of the spring now is √[((L/2)(1-cosθ))2 + (l0 + (L/2)sinθ)2]

Applying the small angle approximation for small displacements, we get

l ≈ l0 + Lθ/2

The magnitude of force exerted by the spring on the rod, F = kLθ/2.

As the angle θ is small, the torque of the force on the rod can be written as

τ = - FL/2 = -kL2θ/4

Now, moment of inertia of the rod about the rotation axis = mL2/12

Therefore, α = τ/I = -3kθ/m = -ω2θ

where ω = √(3k/m)

Thus, time period, T = 2п/ω = 2п√(m/3k)

Putting the values of m and k, we get

T = 2п√(m/3k) = 0.16 s

Q:24 Assume that a narrow tunnel is dug between two diametrically opposite points of the earth. Treat the earth as a solid sphere of uniform density. Show that if a particle is released in this tunnel, it will execute a simple harmonic motion. Calculate the time period of the motion.

Solution: Consider the given figure. Suppose that at a given instant t, the particle in the tunnel is at a distance x from the center of the earth. Only the part of the earth within the sphere of radius x will exert a net attraction on the particle. Mass of this part is

M′ = (4пx3/3)/( 4пR3/3) X M = x3M/R3 where M is the mass of the earth and R is its radius

Therefore, the force of attraction, F = GmM′/x2 = GmMx/R3

This force acts towards the center of the earth. Thus, the resultant force on the particle is opposite to the displacement from the center of the earth and is proportional to it. The particle, therefore, executes a simple harmonic motion in the tunnel with the center of the earth as the mean position.

The force constant is k = GmM/R3, so the time period is

T = 2п√(m/k) = 2п√(R3/GM)

Q:25 All the surfaces shown in figure are frictionless. The mass of the car is M, that of the block is m and the spring has spring constant k. Initially, the car and the block are at rest and the spring is stretched through a length x0 when the system is released. (a) Find the amplitudes of the simple harmonic motion of the block and of the car as seen from the road. (b) Find the time period(s) of the two simple harmonic motions.

Solution: (a) Let the amplitude of oscillation of m and M be x1 and x2.

As there are no external forces involved, the center of mass of the system remains at rest.

Therefore, mx1 = Mx2 …(1)

Also, it is obvious that the block and the car oscillate in opposite directions for conservation of momentum and the center of mass to remain at rest. Hence, the total extension of the spring at the extreme position = x1 + x2.

Therefore, ½ kx02 = ½ k(x1 + x2)2

or, x0 = x1 + x2 …(2)

From equation (1) and (2);

x1 = Mx0/(x1 + x2)

and x2 = mx0/(x1 + x2)

(b) At any position, let v1 and v2 be the velocities of m and M respectively. Here v1 is with respect to M.

As the total energy of the system is constant,

½k(x1 + x2)2 + ½Mv22 + ½m(v1 - v2)2 = constant …(1)

Here v1 - v2 is the absolute velocity of mass ‘m’ as seen from the road.

Again from the law of conservation of momentum,

Mx2 = mx1

or x1 = Mx2/m ….(2)

and Mv2 = m(v1 - v2)

or v1 - v2 = Mv2/m …(3)

Putting (2) and (3) in (1);

½k(Mx2/m + x2)2 + ½Mv22 + ½m(Mv2/m)2 = constant

or ½kx22(M/m + 1)2 + ½Mv22 + ½M2v22/m = constant

or kx22(M/m + 1)2 + Mv22(M/m + 1) = constant

or kx22(M/m + 1) + Mv22 = constant

Differentiating w.r.t. t on both sides;

2kx2(M/m + 1)dx2/dt + 2Mv2 dv2/dt = 0

or 2kx2(M/m + 1)v2 + 2Mv2 a2 = 0

or Ma2 = - kx2(M/m + 1)

or a2 = - kx2((M+m)/Mm)

Thus acceleration is proportional to displacement. Hence the car undergoes simple harmonic motion with time period, T = 2п/ω = 2п√[Mm/(M+m)k]

The time period of the block has to be the same as that of the car because the total momentum and the total displacement of the center of mass of the system has to be zero at all times as there are no external forces involved.

Q:26 In the given figure, a solid cylinder attached to a horizontal spring (k = 5 N/m) rolls without slipping along a horizontal surface. If the system is released from rest when the spring is stretched by 0.3 m, find (a) the translational kinetic energy and (b) the rotational kinetic energy of the cylinder as it passes through the equilibrium position. (c) Show that under these conditions the cylinder’s center of mass executes simple harmonic motion with period

T = 2п√(2M/3k)

where M is the cylinder mass.

Solution: The potential energy at the extreme position is equal to the total kinetic energy (translational + rotational) when it passes through the equilibrium position. Let ‘A’ be the initial extension of the spring. Let ‘v’ and ‘ω’ be the linear velocity and angular velocity of the cylinder respectively when it passes from its equilibrium position. Also, let ‘r’ be the radius of the cylinder.

Therefore, ½kA2 = ½mv2 + ½Iω2

or 5(0.3)2 = mv2 + (½mr2 )(v/r)2

or 0.45 = 3mv2/2

or mv2 = 0.3 J

(a) The translational kinetic energy is, therefore, = ½mv2 = 0.15 J

(b) The rotational kinetic energy = ½Iω2 = mv2/4 = 0.075 J

Therefore, dE/dt = 0

d(3mv2/4)/dt + d(½kx2)/dt = 0

(3mv/2)dv/dt + kxdx/dt = 0

3mva/2 + kxv = 0

a = -2kx/3m

Thus, the acceleration of the cylinder is proportional to its displacement from the equilibrium position. Hence the cylinder executes simple harmonic oscillations with time period,

T = 2п√(3m/2k).

Q:27 For a simple pendulum, find the angular amplitude at which the restoring torque required for simple harmonic motion deviates from the actual restoring torque by 2%. Use g = 10 ms-2.

Solution: The restoring torque on the bob of the pendulum at an angle θ with the vertical is given by τ = -mgsinθ.

We apply small angle approximation to sinθ so that the motion can be followed as simple harmonic motion. The torque becomes τ = -mgθ i.e. it becomes proportional to θ and the motion of the pendulum can now be viewed as simple harmonic motion.

The problem requires us to find the value of θ for which

[(-mgsinθ) - (-mgθ)[/( -mgsinθ) = ±0.02 (we take the absolute value of the expression)

[1 – θ/sinθ] = 0.02

This is an implicit equation and needs to be simplified in irder to be solved for θ.

We expand sinθ using Taylor expansion (upto cubic term as θ is small, so further terms can be neglected.)

sinθ ≈ θ – θ3/6

Therefore, we get

[1 – θ/(θ - θ3/6)] = ±0.02

[1 – 1/(1 – θ2/6)] = ±0.02

Taking the absolute value on both sides, we get

1/(1 – θ2/6) = 0.02 + 1

1/(1 – θ2/6) = 1.02

0.9804 = 1 – θ2/6

θ2/6 = 0.0196

θ = 0.343 rad

or θ = 19.65o

Q:28 In figure, block 2 of mass 2 kg oscillates on the end of a spring in SHM with a period of 20 ms. The block’s position is given by x = (1.0 cm) cos(ωt + п/2). Block 1 of mass 4 kg slides toward block 2 with a velocity of magnitude 6 m/s, directed along the spring’s length. The two blocks undergo a completely inelastic collision at time t = 5 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?

Solution: We know that T = 2п√(k/m) = 25

or 4п2(2/k) = 400 X 10-6

which gives k = 1.97 X 105 N/m

Now, when the collision takes place at t=5 ms, position of block 2 is

x = (1 cm) cos(ωt + п/2) = (1 cm) cos(2пt/T + п/2) = (1 cm) cos(п/2 + п/2) = -1 cm

i.e. block 2 is at its left extreme position and its velocity at that point is zero.

To find the speed of block 2 right after the collision, we use momentum conservation.

We know that after the collision the speed of the two blocks is the same as they stick together because the collision is completely inelastic.

Therefore, m1v1 + m2v2 = (m1+m2)v

or 4 X 6 + 2 X 0 = 6v

which gives v = 4 m/s

Hence, the system now consists of a total mass of 6 kg having kinetic energy, K = ½ mv2

= ½ X 6 X 16 = 48 J

Potential energy of the system is U = ½ kx2 = ½ X 1.97 X 105 X (-0.01)2 ≈ 10 J

Thus the system now has a total mechanical energy of E = K + U = 58 J

When the blocks come to rest at their new extreme position (say X) after the collision, all this energy should be equal to the potential energy of the spring, i.e. ½ kX2 = 58 J

which gives X = 0.024 m = 2.4 cm

Q:29 Hanging from a horizontal beam are nine simple pendulums of the following lengths: (a) 0.10, (b) 0.30, (c) 0.40, (d) 0.80, (e) 1.2, (f) 2.8, (g) 3.5, (h) 5.0, and (i) 6.2 m. Suppose the beam undergoes horizontal oscillations with angular frequencies in the range from 2.00 rad/s to 4.00 rad/s. Which of the pendulums will be (strongly) set in motion? Use g = 10 ms-2.

Solution: We know that ω = √(g/l) for a simple pendulum. The pendulum whose angular frequency matches with the frequency of horizontal oscillations of the rod will be set in motion (strongly).

Using the given values of lengths, we find that the pendulum of length 0.8 m (ω = 3.53 rad/s) and that of length 1.2 m (ω = 2.89 rad/s) will be set in motion as their angular frequency lies in the range of horizontal oscillations of the rod.

Q:30 A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motions, what is the phase difference between the individual motions?

Solution: Let the amplitude of the two simple harmonic motions be ‘A’ and the phase difference between them be ‘ф’.

Both the motions are along the same direction, so x = x1 + x2

x = Asinωt + Asin(ωt + ф)

= Asinωt + Asinωt cosф + Acosωt sinф

= (A + Acosф) sinωt + (Asinф) cosωt

= Csinωt + Dcosωt where C = (A + Acosф) and D = (Asinф)

= √(C2 + D2)[(C/√(C2 + D2))sinωt + (D/√(C2 + D2))cosωt]

Putting C/√(C2 + D2) = cosα and D/√(C2 + D2) = sinα, we get

x = √(C2 + D2)[sinωt cosα + sinα cosωt]

or x = √(C2 + D2) sin(ωt + α)

where the amplitude of the resultant motion is √(C2 + D2) which is equal to A for the given problem. Therefore, A = √(C2 + D2) = √((A + Acosф)2 + (Asinф)2)

or 1 = √(1 + cos2ф + 2cosф + sin2ф)

Using cos2ф + sin2ф = 1;

1 = √(2 + 2cosф)

Now, using cosф = 2cos2(ф/2) – 1, we get

1 = 2cos(ф/2)

or cos(ф/2) = 1/2

which gives ф/2 = 60o

or ф= 120o

Simple Harmonic Motion notes class 11

Contact Form

Advertisement post header

Middle Ad Article 1

Middle Ad Article 2

Article end