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IIT JAM Exam  >  IIT JAM Tests  >  Simple Harmonic Motion MCQ - IIT JAM MCQ

Simple Harmonic Motion MCQ - IIT JAM MCQ


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10 Questions MCQ Test - Simple Harmonic Motion MCQ

Simple Harmonic Motion MCQ for IIT JAM 2024 is part of IIT JAM preparation. The Simple Harmonic Motion MCQ questions and answers have been prepared according to the IIT JAM exam syllabus.The Simple Harmonic Motion MCQ MCQs are made for IIT JAM 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Simple Harmonic Motion MCQ below.
Solutions of Simple Harmonic Motion MCQ questions in English are available as part of our course for IIT JAM & Simple Harmonic Motion MCQ solutions in Hindi for IIT JAM course. Download more important topics, notes, lectures and mock test series for IIT JAM Exam by signing up for free. Attempt Simple Harmonic Motion MCQ | 10 questions in 30 minutes | Mock test for IIT JAM preparation | Free important questions MCQ to study for IIT JAM Exam | Download free PDF with solutions
Simple Harmonic Motion MCQ - Question 1
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The equation of S.H.M of a particle is    is a positive constant. The time period of motion is given by :

Detailed Solution for Simple Harmonic Motion MCQ - Question 1

Compare with equation of SHM of the particle with the standard equation 

Simple Harmonic Motion MCQ - Question 2
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If length of simple pendulum is increased by 6 % then percentage change in time period will be :

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Simple Harmonic Motion MCQ - Question 3
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A particle is executing S.H.M. from mean position at 5 cm distance, acceleration is 20 cm/s2 then value of angular velocity will be :

Detailed Solution for Simple Harmonic Motion MCQ - Question 3

The relation between the acceleration and angular velocity is as follows:
a = ω2x
ω = √a/x
Substituting value of a and x in the above equation:
ω = 2 rad/s.
The correct answer is: 2 rad/s

Simple Harmonic Motion MCQ - Question 4
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A man measures the time period of a simple pendulum inside a stationary lift and finds it to be 10 sec. If the lift accelerates upwards with an acceleration g/4, then find the time period of the pendulum.
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Simple Harmonic Motion MCQ - Question 5
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The maximum velocity of harmonic oscillator is a, and its maximum acceleration is β Its time period will be :
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Simple Harmonic Motion MCQ - Question 6
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One mass m is suspended from a spring. Time period of oscillation is T. Now if spring is divided into n piece and these are joined in parallel order then time period of oscillation if same mass is suspended.
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Simple Harmonic Motion MCQ - Question 7
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Amplitude of harmonic oscillator is a when velocity of particle is half of maximum velocity, then position of particle will be :
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Simple Harmonic Motion MCQ - Question 8
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A simple pendulum suspended from the ceiling of a stationary trolley has a length Its period of oscillation is What will be its period of oscillation if the trolley moves forward with an acceleration f ?
Detailed Solution for Simple Harmonic Motion MCQ - Question 8

The value of time period is when the only force acting on the pendulum is mg its weight and its acceleration is g as shown in the picture below.

Now when the trolly is accelerating with force, f, then the resultant acceleration of the pendulum is 
It is depicted in the diagram below.

Then the resultant time period will be 

Simple Harmonic Motion MCQ - Question 9
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In S.H.M, the graph between kinetic energy 'K' and time 't' is: 
Detailed Solution for Simple Harmonic Motion MCQ - Question 9

The correct answer is:

 

The kinetic energy is KE=  ½mv2,where m is the mass of the pendulum, and v is the speed of the pendulum.

At the highest pointof the pendulum, it is momentarily motionless.  All of the energy in the pendulum is gravitational potential energy and there is no kinetic energy.  At the lowest point the pendulum has its greatest speed.  All of the energy in the pendulum is kinetic energy and there is no gravitational potential energy. This perioic motion continues, hence the graph will be the one shown in the picture.

Simple Harmonic Motion MCQ - Question 10
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Two object A and B of equal mass are suspended from two springs of spring constants KA and KB if the objects oscillate vertically in such a manner that their maximum kinetic energies are equal, then the ratio of their amplitude is :
Detailed Solution for Simple Harmonic Motion MCQ - Question 10

Kinetic energy of A = Kinetic energy of B.

So, ratio of amplitude 
The correct answer is: 

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Simple Harmonic Motion MCQ

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