HBSE Class 11 Physics Simple Harmonic Motion Short Answer Type Questions

Question 21. The oscillation of a body on a smooth horizontal surface is represented by the equation, x = Acosωt, where x = displacement at time t, ω = frequency of oscillation. Which one of the following graphs shows correctly the variation a with t?

Here, a = acceleration at time t and T = time period.

Answer:

x = Acosωt.

So, velocity, \(v=\frac{d x}{d t}=-A \omega \sin \omega t\)

and acceleration, a = \(\frac{d v}{d t}=-A \omega^2 \cos \omega t\)

Now at t = 0, a = -Aω²

and at t = T, a = -Aω²

The option 3 is correct.

Simple Harmonic Motion Class 11 HBSE Notes

Question 22. Two similar springs P and Q have spring constants K_P and K_Q, such that K_P > K_Q. They are stretched, first by the same amount (case a), then by the same force (case b), The work done by the springs W_P and W_Q are related as, in case  1 and case 2, respectively

1. W_P = W_Q; W_P> W_Q
2. W_P = W_Q; W_P= w_Q
3. W_P>W_Q; WQ>W_P
4. W_P< W_Q; W_Q< W_P

Answer:

Work done by the spring (W) = change in potential energy

= \(\frac{1}{2}\) Kx2 (x = Increase in length)

Case 1: The increase in length is the same for P and Q.

Then, \(W_P=\frac{1}{2} K_P x^2 and W_Q=\frac{1}{2} K_Q x^2\)

∴ \(W_P>W_Q\)

Again, if F is the applied force, then the spring constant

K = \(\frac{F}{x} \text { or, } x=\frac{F}{K}\)

∴ W = \(\frac{1}{2} K\left(\frac{F}{K}\right)^2=\frac{1}{2} \frac{F^2}{K}\)

Case 2: \(W_P=\frac{1}{2} \frac{F^2}{K_P} \text { and } W_Q=\frac{1}{2} \frac{F^2}{K_Q}\)

∴ \(W_Q>W_P\)

The option 3 is correct.

Simple Harmonic Motion Definition And Examples Class 11 HBSE

Question 23. When two displacements represented by y₁ = a sinωt and y₂ = bcosωt are superimposed, the motion is

1. Not a simple harmonic
2. Simple harmonic with amplitude a/b
3. Simple harmonic with amplitude \(\sqrt{a^2+b^2}\)
4. Simple harmonic with amplitude \(\frac{(a+b)}{2}\)

Answer:

y = y₁ + y₂ = asinωt + bcosωt

= \(\sqrt{a^2+b^2}\left(\sin \omega t \cdot \frac{a}{\sqrt{a^2+b^2}}+\cos \omega t \cdot \frac{b}{\sqrt{a^2+b^2}}\right)\)

= \(\sqrt{a^2+b^2}(\sin \omega t \cos \theta+\cos \omega t \sin \theta)\)

= \(\sqrt{a^2+b^2} \sin (\omega t+\theta)\)

Hence, this is a simple harmonic motion, whose amplitude \(\sqrt{a^2+b^2}\)

The option 3 is correct.

Question 24. A particle is executing SHM along a straight line. Its velocities at distances x₁ and x₂ from the mean position are V₁ and V₂, respectively. Its time period is

1. \(2 \pi \sqrt{\frac{x_1^2+x_2^2}{V_1^2+V_2^2}}\)
2. \(2 \pi \sqrt{\frac{x_2^2-x_1^2}{V_1^2-V_2^2}}\)
3. \(2 \pi \sqrt{\frac{V_1^2+V_2^2}{x_1^2+x_2^2}}\)
4. \(2 \pi \sqrt{\frac{V_1^2-V_2^2}{x_1^2-x_2^2}}\)

Answer:

⇒ \(V_1=\omega \sqrt{a^2-x_1^2}$ or, $\frac{V_1^2}{\omega^2}=a^2-x_1^2\)

Similarly, \(\frac{V_2^2}{\omega^2}=a^2-x_2^2\)

∴ \(\frac{V_1^2}{\omega^2}-\frac{V_2^2}{\omega^2}=x_2^2-x_1^2 \quad or, \omega=\sqrt{\frac{V_1^2-V_2^2}{x_2^2-x_1^2}}\)

i.e., T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{x_2^2-x_1^2}{V_1^2-V_2^2}}\)

Simple Harmonic Motion Derivation Class 11 HBSE

Question 25. A particle moves so that its position vector is given by \(\vec{r}=\cos \omega t \hat{x}+\sin \omega t \hat{y}\), where co is a constant. Which of the following is true?

1. Velocity and acceleration both are parallel to \(\vec{r}\)
2. Velocity is perpendicular to r and acceleration is directed towards the origin \(\vec{r}\)
3. Velocity is perpendicular to r and acceleration is directed away from the origin \(\vec{r}\)
4. Velocity and acceleration both are perpendicular to \(\vec{r}\)

Answer:

⇒ \(\vec{r}=\cos \omega t \hat{x}+\sin \omega t \hat{y}\)

velocity of the particle, \(\vec{v}=\frac{d \vec{r}}{d t}=-\omega(\sin \omega t) \hat{x}+\omega(\cos \omega t) \hat{y}\)

Acceleration of the particle, \(\vec{a}=\frac{d \vec{v}}{d t}=-\omega^2(\cos \omega t) \hat{x}-\omega^2(\sin \omega t) \hat{y}\)

= \(-\omega^2(\cos \omega t \hat{x}+\sin \omega t \hat{y})=-\omega^2 \vec{r}\)

∴ Acceleration is directed toward the origin.

Again, \(\vec{r}\) . \(\vec{v}\) = -ω sinωt cosωt + ω sinωt cosωt = 0

Thus, \(\vec{v}\) is perpendicular to \(\vec{r}\).

The option 2 is correct.

Question 26. A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s² at a distance of 5 m from the mean position. The time period of oscillation is

1. 2s
2. s
3. 2s
4. 1s

Answer:

⇒ \(\vec{a}=-\omega^2 \vec{x}\)

∴ \(|a|=\omega^2 x \text { or, } 20=\omega^2 \times 5 \text { or, } \omega^2=4 \text { or, } \omega=2\)

T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{2}=\pi \mathrm{s}\)

Option 2 is correct.

Simple Harmonic Motion Formula Class 11 HBSE

Question 27. Explain the relation in phase between displacement, velocity, and acceleration in SHM, graphically as well as theoretically.
Answer:

The relation in phase between displacement, velocity, and acceleration in SHM can be shown graphically.

Displacement x = Asin(ωt + ø₀),

Velocity, v = ωAsin(ωt + ø₀),

Acceleration a = -ω²Asin(ωt + ø₀),

Here,x and v differ in phase by \(\frac{\pi}{2}\), v and a differ in phase by \(\frac{\pi}{2}\).

Types Of Simple Harmonic Motion Class 11 HBSE

Question 28. A particle is in linear simple harmonic motion between two points A and B, 10 cm apart. Take the direction from A to B, as a positive direction and give the signs of velocity and acceleration on the particle when it is

1. At the end of B,
2. At 3 cm away from A going towards B.

Answer:

1. At the end of B, the velocity of the particle is zero. Here acceleration and force both are negative as they are directed along BR i.e., along negative direction.
2. At 3 cm away from A going towards B, the particle is at position R, with a tendency to move along RP which is a positive direction. Here, velocity and acceleration both are positive.

Question 29. The time period of a particle in SHM depends on the force constant k and mass m of the particle as follows: T = \(2 \pi \sqrt{\frac{m}{k}}\). A simple pendulum executes SHM approximately, why then is the time period of a simple pendulum independent of the mass of the pendulum?
Answer:

Given

The time period of a particle in SHM depends on the force constant k and mass m of the particle as follows: T = \(2 \pi \sqrt{\frac{m}{k}}\). A simple pendulum executes SHM approximately,

In case of simple pendulum, the force constant, k = \(\frac{m g}{l}\)

So, time period, T = \(2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{m}{\frac{m g}{l}}}=2 \pi \sqrt{\frac{l}{g}}\)

Hence, T is independent of m.

Characteristics Of Simple Harmonic Motion Class 11

Question 30. One end of the U-tube containing mercury is connected to a suction pump and the other end to the atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Answer:

Given

One end of the U-tube containing mercury is connected to a suction pump and the other end to the atmosphere. A small pressure difference is maintained between the two columns

The restoring force,

F =- weight of mercury column at height 2y

=-(volume) x density x g

=-(A x 2yxρxg) =-2Aρgy

where A = area of cross-section of the tube, ρ = density of mercury.

Force constant k = \(-\frac{F}{y}=2 A \rho \cdot g\)

So, time period T = \(2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{m}{2 A \rho g}}\)

Let L = length of the whole mercury column.

Therefore, mass of mercury, m = volume x density = ALρ

∴ T = \(2 \pi \sqrt{\frac{A L \rho}{2 A \rho g}}=2 \pi \sqrt{\frac{L}{2 g}}\)

where L is the total length of the mercury column =2h, and h is the height of the mercury column in the U-tube. It shows that the mercury column executes SHM.

Question 31. A simple harmonic motion is described by a = -16x where a is acceleration and x is the displacement in meters. What id the time period?
Answer:

Given

A simple harmonic motion is described by a = -16x where a is acceleration and x is the displacement in meters.

Comparing given equation with a = -ω²x, we get

Now, \(\omega=\frac{2 \pi}{T}=\sqrt{16}=4\)

Hence the time period,

T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{4}=\frac{\pi}{2}=\frac{3.14}{2}=1.57 \mathrm{~s}\)

Applications Of Simple Harmonic Motion Class 11 HBSE

Question 32. For a particle executing simple harmonic motion, find the distance from the mean position at which its potential and kinetic energies are equal.
Answer:

Let x₀ = required distance from the mean position

Then, \(\mathrm{KE}=E_x=\frac{1}{2} m \omega^2\left(a^2-x_0^2\right) and \quad \mathrm{PE}=E_p=\frac{1}{2} m \omega^2 x_0^2\)

From the given condition, \(E_k=E_p\)

∴ \(\frac{1}{2} m \omega^2\left(a^2-x_0^2\right)=\frac{1}{2} m \omega^2 x_0^2 or, a^2-x_0^2=x_0^2\)

or, \(2 x_0^2=a^2 \quad or, x_0^2=\frac{a^2}{2} \quad or, x_0= \pm \frac{a}{\sqrt{2}}\)

Question 33. Write the phase difference between the velocity and acceleration of a particle undergoing a simple harmonic motion.
Answer:

In SHM, the phase difference between velocity and acceleration = 90°

Question 33. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed V. If the pendulum makes a small oscillation in a radial direction about its equilibrium position, what will be its time period?
Answer:

Given

A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed V. If the pendulum makes a small oscillation in a radial direction about its equilibrium position

The pendulum bob Is subject to two distinct accelerations:

1. The acceleration due to gravity, \(\overrightarrow{\mathrm{g}}\), acting downwards;
2. The centripetal acceleration, \(\vec{a}=\frac{v^2}{2}\), acting in a horizontal direction. So, the effective, or relative, acceleration due to gravity on the pendulum bob is \(\overrightarrow{g^{\prime}}=\vec{g}-\vec{a}=\vec{g}+(\overrightarrow{-a})\)

Its magnitude, \(g^{\prime}=\sqrt{g^2+\frac{V^{\prime}}{R^2}} \text {. }\)

Then, the time period of the pendulum becomes \(T^{\prime}=2 \pi \sqrt{\frac{l}{g^{\prime}}}=\frac{2 \pi \sqrt{l}}{\left(g^2+\frac{V^4}{R^2}\right)^{1 / 4}}\)

Difference Between SHM And Oscillatory Motion Class 11

Question 34. What will be the change in the time period of a loaded spring, when taken to the moon?
Answer:

Time period, \(T \propto \frac{1}{\sqrt{g}} .\). As g, the acceleration due to gravity is less on the surface of the moon, the time period would increase.

Question 35. Show the variation of kinetic energy, potential energy, and total energy with

1. Time and
2. Displacement graphically.

Answer:

The displacement is x = acosωt; this means that the motion starts at one end of the path, comes to the mean position at \(\frac{T}{4}\) (T = time period), goes to the other end at \(\frac{T}{2}\), then again comes to the mean position at \(\frac{3T}{4}\), and finally comes back to the starting point at time T, thus completing a complete oscillation.

The variation of the KE, the PE, and the total energy

1. With time t is shown in the graph
2. With displacement, x is shown in the graph

Question 36. How will the time period of a simple pendulum change when its length is doubled?
Answer:

Time period of a simple pendulum, T = \(2 \pi \sqrt{\frac{l}{g}}\)

So for two different lengths, \(\frac{T_1}{T_2}=\sqrt{\frac{l_1}{L_2}}\)

or, \(T_2=T_1 \sqrt{\frac{l_2}{l_1}}=T_1 \sqrt{\frac{2}{1}}=\sqrt{2} T_1\) (given, \(l_2=2 l_1\))

Hence, the time period becomes √2 times when the length is doubled.

Energy In Simple Harmonic Motion Class 11 HBSE

Question 36. Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown. Show that the mass executes simple harmonic motion when displaced from its rest position on either side.

Answer:

Let the block of mass be displaced by a small distance x to the right side of the equilibrium position as shown. Under this situation the spring on the left side of the mass gets elongated by a length equal to x and that on the right side gets compressed by the same length

The forces acting on the mass are F₁ = -kx

(Exerted by the spring on the left side, trying to pull the mass towards the mean position)

F₂ = -kx

(exerted by the spring on the right side, trying to push the mass towards the mean position)

Hence, the net force acting on the block of mass is F = -2kx which is proportional to the displacement x and is directed towards the mean position. Therefore, the motion executed by the block of mass is simple harmonic.

Question 37. y(t) = (sinωt- cosωt) represents simple harmonic motion, determining the period of SHM.
Answer:

Given

y(t) = (sinωt- cosωt)

= \(\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin \omega t-\frac{1}{\sqrt{2}} \cos \omega t\right)\)

= \(\sqrt{2}\left(\sin \omega t \cos \frac{\pi}{4}-\cos \omega t \sin \frac{\pi}{4}\right)=\sqrt{2} \sin \left(\omega t-\frac{\pi}{4}\right)\)

It is an equation of SHM. Where angular frequency is and time period T = \(\frac{2 \pi}{\omega}\)

Numerical Problems On SHM Class 11 HBSE

Question 38. At what displacement

1. The PE and
2. KE of a simple harmonic oscillator is maximum?

Answer:

1. PE of a simple harmonic oscillator is maximum at x = ±A

PE at that instant, U = \(\frac{1}{2} m \omega^2 A^2\)

KE of a simple harmonic oscillator is maximum at the position of equilibrium, x = 0.

KE at that instant, K = \(\frac{1}{2} m \omega^2 A^2\)