Test: Simple Harmonic Motion


10 Questions MCQ Test Physics Class 11 | Test: Simple Harmonic Motion


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QUESTION: 1

The velocity of a particle moving with simple harmonic motion is____ at the mean position.

Solution:

Equation of SHM particle: 
Y=a sinωt
V=aω sinωt
Vmax = aω
So the velocity is maximum at mean position

QUESTION: 2

The periodic time (tp) is given by:

Solution:

Periodic time is the time taken for one complete revolution of the particle.

∴ Periodic time (tp) = 2 π/ω seconds.

QUESTION: 3

A frequency of 1Hz corresponds to:

Solution:

Frequency is defined as time taken to perform one oscillation by the object.

Hence, 1Hz corresponds to 1 vibration per sec.

QUESTION: 4

A second pendulum is mounted in a space shuttle. Its period of oscillations will decrease when rocket is:

Solution:

 

  • Time Period, T = 2π √(l/g')where,
    l = Length of seconds pendulum 
    g’ = Apparent Gravity
  • For the period of oscillations of Seconds Pendulum to decrease, the Apparent gravity (g’) has to increase because:
  • Hence, Time Period of oscillations of Seconds Pendulum will decrease when the rocket is ascending up with uniform acceleration.
QUESTION: 5

A particle of mass 10 gm lies in a potential field v = (50x2+100) J/kg. The value of frequency of oscillations in Hz is

Solution:

QUESTION: 6

If a simple pendulum oscillates with an amplitude 50 mm and time period 2s, then its maximum velocity is:

Solution:

We know that in a simple harmonic motion the maximum velocity,
Vmax = A⍵
Here A = 50 mm

And ⍵ = 2π / T
= 2π / 2
= π

Hence  Vmax = 50 x 10-3
= 0.15 m/s

QUESTION: 7

 In simple harmonic motion the displacement of a particle from its equilibrium position is given by 11915_image018. Here the phase of motion is 

Solution:
QUESTION: 8

Find the amplitude of the S.H.M whose displacement y in cm is given by equation y= 3 sin157t + 4 cos157t, where t is time in seconds.

Solution:

When the displacement of a SHM is:
y=a sin wt+ b cos wt

  • Amplitude of the SHM will be:
    A=√a2+b2

Here, a = 3, b = 4
Amplitude, A= √(32+42) = 5 cm

Hence option B is correct.

QUESTION: 9

What will be the phase difference between bigger pendulum (with time period 5T/4 ) and smaller pendulum (with time period T) after one oscillation of bigger pendulum?

Solution:

After one oscillation of a bigger pendulum i.e. 5T/4, ¼ of the total phase is travelled by the smaller pendulum while the bigger is still at initial position.

Thus, the phase difference between two is: ¼ (2π) - 0 = π/2

QUESTION: 10

A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:

Solution:

   ∴ We get, ω = √3 s-1
   T = 2π / √3

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