Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Solve this problem A particle at the end of a spring executes simple harmonic motion with a period t1 , while the corresponding period for another spring is t2 . If the period of oscillation with the two springs in series is T, then

A particle at the end of a spring executes simple harmonic motion with a period t1 , while the corresponding period for another spring is t2 . If the period of oscillation with the two springs in series is T, then

  • Option 1)

    T=t_{1}+t_{2}\; \; \;

  • Option 2)

    T^{2}=t_{1}^{2}+t_{2}^{2}\; \;\; \;

  • Option 3)

    T^{-1}=t_{1}^{-1}+t_{2}^{-1}\; \; \; \;

  • Option 4)

    T^{-2}=t_{1}^{-2}+t_{2}^{-2}

 

Answers (1)

best_answer

As we learnt in

Series combination of spring -

- wherein

\frac{1}{K_{eq}}= \frac{1}{K_{1}}+ \frac{1}{K_{2}}

K_{1}and\ K_{2} are spring constants of spring 1 & 2 respectively.

 

 

 

When springs are in series, k=\frac{k_{1}k_{2}}{k_{1}+k_{2}}

For first spring,  t_{1}=2\pi \sqrt{\frac{m}{k_{1}}}

For second spring  t_{2}=2\pi \sqrt{\frac{m}{k_{2}}}

\therefore \; \; \; t_{1}^{2}+t_{2}^{2}=\frac{4\pi ^{2}m}{k_{1}}+\frac{4\pi ^{2}m}{k_{2}}=4\pi ^{2}m\left ( \frac{k_{1}+k_{2}}{k_{1}k_{2}} \right )

or \; \; \; t_{1}^{2}+t_{2}^{2}=\left [ 2\pi \sqrt{\frac{m(k_{1}+k_{2})}{k_{1}k_{2}}} \right ]^{2}

or \; \; \; t_{1}^{2}+t_{2}^{2}=T^{2}

Correct option is 2.


Option 1)

T=t_{1}+t_{2}\; \; \;

This is an incorrect option.

Option 2)

T^{2}=t_{1}^{2}+t_{2}^{2}\; \;\; \;

This is the correct option.

Option 3)

T^{-1}=t_{1}^{-1}+t_{2}^{-1}\; \; \; \;

This is an incorrect option.

Option 4)

T^{-2}=t_{1}^{-2}+t_{2}^{-2}

This is an incorrect option.

Posted by

Plabita

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 2025 Registration Live: NTA session 1 application last date today; official website, syllabus
anam-khan

JEE Main 2025 session 1 registration ends today at jeemain.nta.nic.in; application correction from November 26
anam-khan

JEE Main 2025 registration ends tomorrow; changes introduced in exam this year

Latest Question