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Simple Harmonic Motion NAT Level - 1


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10 Questions MCQ Test Topic wise Tests for IIT JAM Physics | Simple Harmonic Motion NAT Level - 1

Simple Harmonic Motion NAT Level - 1 for IIT JAM 2023 is part of Topic wise Tests for IIT JAM Physics preparation. The Simple Harmonic Motion NAT Level - 1 questions and answers have been prepared according to the IIT JAM exam syllabus.The Simple Harmonic Motion NAT Level - 1 MCQs are made for IIT JAM 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Simple Harmonic Motion NAT Level - 1 below.
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*Answer can only contain numeric values
Simple Harmonic Motion NAT Level - 1 - Question 1

A spring has a certain mass suspended from it and its period for vertical oscillation is T1. The spring is now cut into two equal halves and the same mass is suspended from one of the half. The period of vertical oscillation is now T2. The ratio of   is :


Detailed Solution for Simple Harmonic Motion NAT Level - 1 - Question 1


When spring is cut in equal part, then force constant is 

The correct answer is: 0.707

*Answer can only contain numeric values
Simple Harmonic Motion NAT Level - 1 - Question 2

For a simple harmonic vibrator of frequency n, the frequency of kinetic energy changing completely to potential energy is αn. Find the value of α.


Detailed Solution for Simple Harmonic Motion NAT Level - 1 - Question 2

2n
The correct answer is: 2

*Answer can only contain numeric values
Simple Harmonic Motion NAT Level - 1 - Question 3

The total energy of the body executing S.H.M is E. Then the kinetic energy when the displacement is half of the amplitude is αE Find the value of α.


Detailed Solution for Simple Harmonic Motion NAT Level - 1 - Question 3

Kinetic energy  
Kinetic energy  
The correct answer is: 0.75

*Answer can only contain numeric values
Simple Harmonic Motion NAT Level - 1 - Question 4

The amplitude of a damped harmonic oscillator become halved in 1 minute. After three minutes the amplitude will become 1/x of initial amplitude where x is :


Detailed Solution for Simple Harmonic Motion NAT Level - 1 - Question 4

The variation in amplitude of a damped harmonic oscillation with time is given by

 initial amplitude,
 damped factor
It is given that after 1 minutes  

The correct answer is: 8

*Answer can only contain numeric values
Simple Harmonic Motion NAT Level - 1 - Question 5

One body of mass m is suspended from three springs as shown in figure each spring has spring constant k. If mass m is displaced slightly then time period of oscillation is  Find the value of  α.


Detailed Solution for Simple Harmonic Motion NAT Level - 1 - Question 5

Using series and parallel combination of spring

The correct answer is: 1.5

*Answer can only contain numeric values
Simple Harmonic Motion NAT Level - 1 - Question 6

A particle is executing SHM with an amplitude 4 cm. The displacement (in cm) at which its energy it half kinetic and half potential is :


Detailed Solution for Simple Harmonic Motion NAT Level - 1 - Question 6


Potential energy = Kinetic energy

The correct answer is: 2.828

*Answer can only contain numeric values
Simple Harmonic Motion NAT Level - 1 - Question 7

A particle executing S.H.M of amplitude 4 cm and T = 4 sec. The time taken (in sec) by it to move from positive extreme position to half the amplitude is :


Detailed Solution for Simple Harmonic Motion NAT Level - 1 - Question 7


The correct answer is: 0.667

*Answer can only contain numeric values
Simple Harmonic Motion NAT Level - 1 - Question 8

The potential energy of a particle executing S.H.M. is 2.5 J when its displacement is half of amplitude. The total energy (in Joule) of the particle is :


Detailed Solution for Simple Harmonic Motion NAT Level - 1 - Question 8

Total energy  
Potential energy 

So total energy 
The correct answer is: 10

*Answer can only contain numeric values
Simple Harmonic Motion NAT Level - 1 - Question 9

If the time period of oscillation of mass M suspended from a spring is one second, then the time period (in sec) of 4M will be :


Detailed Solution for Simple Harmonic Motion NAT Level - 1 - Question 9


The correct answer is: 2

*Answer can only contain numeric values
Simple Harmonic Motion NAT Level - 1 - Question 10

A linear harmonic oscillation of force constant 2 x 106 Nlm and amplitude 0.01 m has a total mechanical energy of 160 joules. Its maximum K.E. (in Joule) will be.


Detailed Solution for Simple Harmonic Motion NAT Level - 1 - Question 10

Maximum kinetic energy is 100 J
Maximum kinetic energy 
The correct answer is: 100

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