Simple Harmonic Motion Test

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Simple Harmonic Motion Test - 17

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Question 1
1 / -0

A particle performs S.H.M. with frequency f. Frequency of its velocity and acceleration are respectively :

A
f, f

B
f/2, f

C
2f, f

D
f, 2f

Solution

The frequency of velocity and acceleration is equal to the frequency of oscillation.
Question 2
1 / -0

Force acting on a body free to move on X-axis is given by F = −kxⁿ where k is a positive constant. For which value of n motion of the body is not oscillatory?

A
3

B
7

C
2

D
5

Solution

For oscillatory motion,

F α -x

For n=2, when x is -ve, force is also -ve. So, it is not opposite to the displacement.
Question 3
1 / -0

A particle is executing S.H.M. Graph of its displacement with the position is shown in the figure. Maximum acceleration of the particle is

A

B

C

D

Solution

From the graph: A = 2 m, T = 4 sec
Question 4
1 / -0

A particle is executing SHM along a straight-line path. Then which of the following statement is correct?

A
Acceleration always decreases the speed of the particle

B
Acceleration is constant with time

C
Acceleration is constant with position

D
Acceleration is maximum when the particle is at rest

Solution

Acceleration of the particle is maximum when the particle is at the extreme position and the particle is at rest at the extreme position.
Question 5
1 / -0

A particle is moving along X-axis (x in meter) and force acting on particle varies as F = -5x + 5N, then

A
The motion of particle will be circular

B
The motion of particle will be periodic having a mean position at x = 1m

C
The equilibrium position of the particle will be at x = 4m

D
Both (1) & (2)

Solution
Question 6
1 / -0

The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:

A

B

C
Zero

D
π rad

Solution

Hint: Find the equations of displacement and acceleration.

Step 1: Write the equations of displacement and acceleration.

x = Asinωt

a = Asin(ωt+π)

Step 2: Find the phase difference.

The phase difference, ϕ = π.
Question 7
1 / -0

Which of the following examples represent simple harmonic motion?

A
The rotation of the earth about its axis.

B
The motion of an oscillating mercury column in a U-tube.

C
General vibrations of a polyatomic molecule about its equilibrium position.

D
A fan rotating with a constant angular velocity.

Solution

Hint: For a simple harmonic motion, a = −kx

Step 1: Find the pressure difference created.

Let us assume that the length of the liquid column in each arm of the u-tube is in equilibrium to be 'l'. Also, let 'ρ' be the density of the mercury in the u-tube and 'A' be the area of the cross-section of the liquid column.

If the distance pushed by mercury in one arm of the u-tube is 'x' then the level of mercury in the other arm will be raised by 'x'.

The pressure difference experienced due to the difference in the level of mercury in u tube will be,

ΔP=ρgx

Step 3: Find the relation between acceleration and displacement.

Force due to pressure,

ma = −PA

ma = −ρgxA

a ∝ − x
Question 8
1 / -0

The period of ( sinωt−cos ωt) is:

A
π/ω

B
4π/ω

C
3π/ω

D
2π/ω

Solution

Hint: Use the concept of superposition of the oscillations.

Step 1: Combine both of the oscillations.

Given, the equation is,

y = sinωt − cosωt

= sinωt − sin(90° + ωt )

Step 2: Find the time period of the oscillation.

Angular frequency = ω

Time period = 2π/ω
Question 9
1 / -0

y = sin³ωt is

A
a simple harmonic function.

B
a simple harmonic function but not a periodic function.

C
both periodic and simple harmonic function.

D
only periodic function.

Solution

Hint: In case of SHM, the force is directly proportional to the displacement.

Step 1: Modify the equation.

y = sin³ωt

Using,

As both are periodic, y is also a periodic function.

Step 2: Find the force equation.

On differentiating y, w.r.to t:

As acceleration is not directly proportional to y, it is not an SHM.
Question 10
1 / -0

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 the positive direction.) The signs of velocity, acceleration, and force on the particle when it is at 3 cm away from A going towards B are:

A
Zero, Negative, Negative

B
Negative, Zero, Zero

C
Negative, Negative, Negative

D
Positive, Positive, Positive

Solution

Hint: The position of the particle w.r.to the mean position defines the direction of velocity and acceleration.

Step 1: Find the direction of velocity and acceleration.

The velocity is directed from A to B, so the velocity is positive.

The particle experiences a force towards the mean position (O) in the direction AB, so the force is positive.

The force and acceleration will always have the same signs, so the acceleration is positive.