Question

A spring has a certain mass suspended from it and period for vertical oscillation is T . The spring is now cut into two equal halves and the same mass is suspended from one of the halves.The period of vertical oscillation is now..​

Answer 1

Answer:

Given:

Initial time period = T

Spring is cut into 2 pieces and again suspended from one half

To find:

New time period

Calculation:

Let the initial Spring constant be k and final spring constant be k2 . Since it is cut, we can say that :

$= > k \times l = (k2) \times \dfrac{l}{2}$

$= > k = \dfrac{(k2)}{2}$

$= > k2 = 2k$

Initial time period = T

$= > T = 2\pi \sqrt{ \dfrac{m}{k} }$

Putting new constant for spring :

$T2 = 2\pi \sqrt{ \dfrac{m}{(k2)} }$

$= > T2 = 2\pi \sqrt{ \dfrac{m}{(2k)} }$

$= > T2 = \dfrac{1}{ \sqrt{2} } \bigg(2\pi \sqrt{ \dfrac{m}{k} } \bigg)$

$= > T2 = \dfrac{T}{ \sqrt{2} }$

So final answer :

$\boxed{ \blue{ \huge{ \bold{ T2 = \dfrac{T}{ \sqrt{2} } }}}}$

Answer 2

$\tt{\huge{\pink{Hello!!! }}}$

$</p><p>\tt{\blue{\implies{ k = \dfrac{k_{2}}{2} }}}$

$\tt{\red{\implies{k_{2} = 2k }}}$

Initial time period = T. Hence,

$\tt{\orange{\implies{ T = 2 \pi \sqrt{ \dfrac{m}{k} }}}}$

$\tt{\green{\implies{T_{2} = 2 \pi \sqrt{ \dfrac{m}{K_{2} }}}}}$

$\tt{\red{\implies{ T2 = \dfrac{1}{ \sqrt{2} } \bigg(2\pi \sqrt{ \dfrac{m}{k} } \bigg) }}}$

$</p><p></p><p> \tt{\blue{\implies{ T_{2} = \dfrac{T}{ \sqrt{2} }}}}$