Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Help me understand! - Oscillations and Waves - JEE Main-6

A particle moves with simple harmonic motion in a straight line. In first \tau s, after starting from rest it travels a distance a, and in next  \tau s it travels 2a, in same direction, then :

  • Option 1)

    amplitude of motion is 3a
     

     

  • Option 2)

     time period of oscillations is 8\pi

  • Option 3)

     amplitude of motion is 4a

     

  • Option 4)

     time period of oscillations is 6\pi

 

Answers (1)

As we learned

 

Equation of S.H.M. -

a=-\frac{d^{2}x}{dt^{2}}= -w^{2}x

w= \sqrt{\frac{k}{m}}

 

- wherein

x= A\sin \left ( wt+\delta \right )

 

 

x=A\cos wt

a+t=0_{1}, x=A

a+t=\tau _{1}

i+ravesed Q

\Rightarrow A-a=A\cdot \cos wt

At  t=2t, the particle traveled 3Q

\Rightarrow A-3a=A\cos2wt

as \cos2wt =2\cos ^{2}wt-1

\Rightarrow \frac{A-3a}{A}=2\cdot \left ( \frac{A-a}{A} \right )^{2}-1

\Rightarrow a^{2}=2aA  or A=2a

Now A-a = Acoswt

or cos_{wt}\cdot \frac{1}{2} or

 

\frac{2\pi }{\tau }\cdot \tau =\frac{\pi }{3}\Rightarrow \tau =6\tau

 


Option 1)

amplitude of motion is 3a
 

 

Option 2)

 time period of oscillations is 8\pi

Option 3)

 amplitude of motion is 4a

 

Option 4)

 time period of oscillations is 6\pi

Posted by

Vakul

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 session 2 result date announced; Steps to check NTA session 2 scores
anam-khan

JEE Main 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25
anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today

Latest Question