Simple Harmonic Motion Test

Result

Simple Harmonic Motion Test - 4

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Question 1
1 / -0

The variation of acceleration of a particle executing SHM with displacement x is:

A

B

C

D

Solution

The acceleration of a particle performing S.H.M. is given by this expression:

a = −ω²x
Question 2
1 / -0

The motion of a particle varies with time according to the relation y = a sin ωt + a cos ωt. Then

A
The motion is oscillatory but not SHM

B
The motion is SHM with an amplitude a√2

C
The motion is SHM with an amplitude √2

D
The motion is SHM with an amplitude a

Solution
Question 3
1 / -0

If a particle is executing SHM, with an amplitude A, the distance moved and the displacement of the body in a time equal to its time period are, respectively,

A
2A, A

B
4A, 0

C
A, A

D
0, 2A

Solution

The distance travelled by the particle in one time period is 4A and displacement = 0.
Question 4
1 / -0

The equations of the displacement of two particles making SHM are represented by y₁ = a sin (ωt + φ) and y₂ = a cos (ωt) respectively. The phase difference of the velocities of the two particles will be

A
π/2 + φ

B
-φ

C
φ

D
φ - π/2

Solution
Question 5
1 / -0

The displacement of a particle executing SHM is given by y = 0.25 (sin 200t) cm. The maximum speed of the particles is:

A
200 cm/sec

B
100 cm/sec

C
50 cm/sec

D
0.25 cm/sec

Solution
Question 6
1 / -0

A particle is executing SHM with an amplitude A and the time period T. If at t = 0, the particle is at origin (mean position), then the time instant when it covers a distance equal to 2.5A will be

A
T/12

B
5T/12

C
7T/12

D
2T/3

Solution

The equation of SHM: x = A sin ωt
Question 7
1 / -0

A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude?

A
1/2 s

B
1/6 s

C
1/4 s

D
1/3 s

Solution
Question 8
1 / -0

The uniform stick of mass m length L is pivoted at the centre. In the equilibrium position shown in the figure, the identical light springs have their natural length. If the stick is turned through a small angle θ, it executes SHM. The frequency of the motion is:

A

B

C

D
None of these

Solution

If the linear velocity of the end of the rod is v at any instant,
Question 9
1 / -0

If the displacement x and the velocity v of a particle executing simple harmonic motion are related through the expression 4v2=25−x24v2=25-x2,then its time period will be

A
π

B
2π

C
4π

D
6π

Solution
Question 10
1 / -0

A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector is correctly shown i

A

B

C

D

Solution

a_c = Centripetal acceleration

a_t = tangential acceleration

a_N = net acceleration = Resultant of a_c and a_t
Question 11
1 / -0

A second's pendulum is mounted in a rocket. Its period of oscillation decreases when the rocket:

A
Comes down with uniform acceleration

B
Moves around the earth in a geostationary orbit

C
Moves up with a uniform velocity

D
Moves up with the uniform acceleration

Solution

It is possible when rocket moves up with uniform acceleration a. Then g_eff = g + a