Simple Harmonic Motion Test

Result

Simple Harmonic Motion Test - 6

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Question 1
1 / -0

An S.H.M. has an amplitude ‘a’ and a time period T. The maximum velocity will be -

A
4a/T

B
2a/T

C
2π/T

D
2aπ/T

Solution
Question 2
1 / -0

Two particles P and Q start from origin and execute Simple Harmonic Motion along X-axis with same amplitude but with periods 3 seconds and 6 seconds respectively. The ratio of the velocities of P and Q when they meet is -

A
1 : 2

B
2 : 1

C
2 : 3

D
3 : 2

Solution

The particles will meet at the mean position when P completes one oscillation and Q completes half an oscillation
Question 3
1 / -0

The angular velocities of three bodies in simple harmonic motion are ω₁,ω_2,ω₃ with their respective amplitudes as A_1,A₂,A₃. If all the three bodies have same mass and maximum velocity, then

A
A₁ω₁₌A₂ω₂=A₃ω₃

B
A₁ω₁²= A₂ω₂²= A₃A₃²

C
A₁²ω₁=A₂²ω₂=A₃²ω₃

D
A₁²ω₁²= A₂²ω₂²= A²

Solution

Velocity is the same. So by using-

A₁ω₁= A₂ω₂= A₃A₃
Question 4
1 / -0

The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes 8√3 cm/sec will be

A
2√3 cm

B
√3 cm

C
1 cm

D
2 cm

Solution

At mean position velocity is maximum
Question 5
1 / -0

The maximum velocity of simple harmonic motion represented by y = 3sin (100t+6/π) is given by

A
300

B
3π/6

C
100

D
π/6

Solution

From the equation of the wave,

Amplitude of wave is A = 3

The angular velocity of the wave can be obtained as ω=100 rad/sec

We know that the velocity can be obtained as:

v_max = aω

= 3×100 = 300
Question 6
1 / -0

The displacement equation of a particle is x=3 sin 2t + 4cos 2t The amplitude and maximum velocity will be respectively

A
5, 10

B
3, 2

C
4, 2

D
3, 4

Solution

x = 3 sin 2t + 4 cos 2t From given equation

a₁ = 3, a₂= 4 and ∅ = π/2
Question 7
1 / -0

The instantaneous displacement of a simple pendulum oscillator is given by x = A cos (ωt + π/4) . Its speed will be maximum at time

A
π/4ω

B
π/2ω

C
π/ω

D
2π/ω

Solution
Question 8
1 / -0

The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is

A
144π²m/sec²

B
144m/sec²

C

D
288π²m/sec²

Solution

ω = 2πn where n is frequency

Maximum acceleration

= 0.01 × 4 ×(π)²× (60)²= 144π²m/sec²
Question 9
1 / -0

A particle moving along the x-axis executes simple harmonic motion, then the force acting on it is given by

A
– A Kx

B
A cos (Kx)

C
A exp (– Kx)

D
A Kx

Solution

For S.H.M. F = − kx

so Force = Mass × Acceleration ∝ − x

⇒ F = – Akx; where A and k are positive constants
Question 10
1 / -0

What is the maximum acceleration of the particle doing the SHM where 2 is in cm

A
π/2 cm/s²

B
π²/2 cm/s²

C
π/4 cm/s²

D
π/4 cm/s²

Solution

Comparing given equation with standard equation