Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A uniform spring whose initial length is $l$ has force const $K$ . Spring is cut into 2 pieces of length $l_{1}$ & $l_{2}$ such that $l_{1} = nl_{2}$ , $n$ is integer. Now mas ' $m$ ' is attached to first spring. The time period of its oscillation will be

Option 1)
$T = 2\pi \sqrt{\frac{mn}{K(n+1)}}$

Option 2)
$T = 2\pi \sqrt{\frac{m}{K(n+1)}}$

Option 3)
$T = 2\pi \sqrt{\frac{m}{nK}}$

Option 4)
$T = 2\pi \sqrt{\frac{m(n+1)}{nK}}$

Answers (1)

best_answer

$K\alpha \frac{1}{l} \Rightarrow K' = \left ( \frac{n+1}{n} \right )K$

$\Rightarrow T = 2\pi\sqrt{\frac{mn}{K(n+1)}}$

Cutting of Spring -

If a spring is cut into 2 peices of length $l_{1}$ & $l_{2}$ where $l_{1} = nl_{2}$

Spring const of

* First part $K_{1} = \frac{K(n+1)}{n}$

$K_{2} = (n+1)K$

- wherein

Formula

$\frac{K_{1}}{K_{2}} = \frac{1}{n}$

Option 1)

$T = 2\pi \sqrt{\frac{mn}{K(n+1)}}$

This is correct.

Option 2)

$T = 2\pi \sqrt{\frac{m}{K(n+1)}}$

This is incorrect.

Option 3)

$T = 2\pi \sqrt{\frac{m}{nK}}$

This is incorrect.

Option 4)

$T = 2\pi \sqrt{\frac{m(n+1)}{nK}}$

This is incorrect.

Posted by

prateek

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: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective

anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’

anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile

Latest Question