Simple Harmonic Motion Test

Result

Simple Harmonic Motion Test - 16

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Question 1
1 / -0

Equation of simple harmonic motion of a particle is y = (0.4 m) sin314t, where time t is in second. Frequency of vibration of the particle is

A
100 Hz

B
75 Hz

C
50 Hz

D
25 Hz

Solution
Question 2
1 / -0

A particle starts SHM from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of its maximum speed, its displacement is

A
A/2

B
A/√2

C

D

Solution
Question 3
1 / -0

The acceleration of a particle in SHM is

A
Always zero

B
Always constant

C
Maximum at the extreme position

D
Maximum at the equilibrium position

Solution

Acceleration is given by -

a = −ω²x

It is maximum when x is maximum i.e. at the extreme position.
Question 4
1 / -0

A particle moves in the XY plane according to the equation The motion of the particle is along:

A
Straight line and periodic

B
Circle and non-periodic

C
Ellipse and periodic

D
Parabola and non-periodic

Solution

It is a straight line and the motion of the particle is periodic as it is a form of the sine function.
Question 5
1 / -0

A particle is performing SHM with amplitude A and angular velocity ω. The ratio of the magnitude of maximum velocity to maximum acceleration is

A
ω

B
1/ω

C
ω²

D
Aω

Solution
Question 6
1 / -0

Values of the acceleration A of a particle moving in simple harmonic motion as a function of its displacement x are given below

The period of the motion is :

A

B

C

D
πs

Solution
Question 7
1 / -0

The acceleration-time graph of a particle undergoing SHM is shown in the figure.

A
The velocity of the particle at point 2 is zero

B
Velocity at point 3 is zero

C
Velocity at point 2 is +ve and maximum

D
Both (2) & (3)

Solution

The particle is at the mean position at the point (2) and is at the extreme position at the point (3).

At the point (2), the particle is moving towards +x-direction as acceleration is becoming -ve.
Question 8
1 / -0

A particle is executing SHM about origin along X-axis, between points A(α, 0) and B(-α, 0). Its time period of oscillation is T. The magnitude of its acceleration T/12 second after the particle reaches point A will be

A

B

C

D

Solution
Question 9
1 / -0

A particle executes linear oscillation such that its epoch is zero. The ratio of the magnitude of its displacement in 1st second and 2nd second is (Time period = 12 seconds)

A

B

C

D

Solution
Question 10
1 / -0

A simple pendulum attached to the ceiling of a stationary lift has a time period of 1 s. The distance y covered by the lift moving downward varies with time as y = 3.75 t², where y is in meters and t is in seconds. If g = 10 m/s², then the time period of the pendulum will be

A
4 s

B
6 s

C
2 s

D
12 s

Solution