Simple Harmonic Motion Test

Result

Simple Harmonic Motion Test - 7

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Question 1
1 / -0

The equation of motion of a particle is where K is positive constant. The time period of the motion is given by

A
2π/K

B
2πK

C

D
2π/√K

Solution

On comparing with standard equation
Question 2
1 / -0

A simple harmonic wave having an amplitude a and time period T is represented by the equation y = 5 sinπ(t+4)m Then the value of amplitude (a) in (m) and time period (T) in second are

A
a=10, T=2

B
a=5, T=1

C
a=10, T=1

D
a=5, T=2

Solution

y = 5sin (πt + 4π), comparing it with standard equation
Question 3
1 / -0

The period of a simple pendulum measured inside a stationary lift is found to be T. If the lift starts accelerating upwards with acceleration of g/3 then the time period of the pendulum is

A
T/√3

B
T/3

C
√3/2 T

D
√3T

Solution
Question 4
1 / -0

The time period of a simple pendulum of length L as measured in an elevator descending with acceleration g/3 is

A

B

C

D

Solution

The effective acceleration in a lift descending with acceleration
Question 5
1 / -0

If a body is released into a tunnel dug across the diameter of earth, it executes simple harmonic motion with time period

A

B

C

D

Solution

Acceleration due to gravity at a depth d below the surface of the earth;
Question 6
1 / -0

If the displacement equation of a particle be represented by y = Asin Pt + B cos Pt, the particle executes

A
A uniform circular motion

B
A uniform elliptical motion

C
A S.H.M.

D
A rectilinear motion

Solution

y = A sin Pt + B cos Pt

let A = r cosθ, B = r sinθ

⇒ y = r sin (Pt + θ) which is the equation of SHM.
Question 7
1 / -0

The displacement of a particle varies according to the relation x = 4(cosπt + sinπt). The amplitude of the particle is

A
8

B
– 4

C
4

D
4√2

Solution

For given relation
Question 8
1 / -0

A S.H.M. is represented by x = 5√2 (sin 2πt + cos 2πt). The amplitude of the S.H.M. is

A
10 cm

B
20 cm

C
5√2 cm

D
50 cm

Solution
Question 9
1 / -0

The displacement of a particle varies with time as x = 12 sin wt − 16 sin³ wt (in cm). If its motion is S.H.M., then its maximum acceleration is -

A
12ω²

B
36ω²

C
144ω²

D
√192ω²

Solution

Acceleration is maximum when x = A

So maximum acceleration: a_max= (3ω)²× 4 = 36ω²
Question 10
1 / -0

A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is U (x) = k [x]³ , where k is a positive constant. If the amplitude of oscillation is a, then its time period T is -

A
Proportional to 1√a

B
Independent of a

C
Proportional to √a

D
Proportional to a^3/2

Solution