Question

A spring of force constant K is attached with a mass executes with S.H.M has a time period T. It is cut into three equal parts. If the springs are connected in parallel then the time period of the combination for the same mass is

• A. T
• B. T/3
• C. T/9
• D. 2T/3

Answer 1

The correct option is B T/3

A spring of length $l$ can be considered as a series combination of spring of length $\frac{l}{3}$ and the equivalent spring constant of the original spring. Now, if $k^{\prime }$ is the spring constant of each of the three spring then $k$ is the spring constant of original spring, then $\frac{1}{k}=\frac{1}{k^{\prime }}+\frac{1}{k^{\prime }}+\frac{1}{k^{\prime }}$

$\Rightarrow k=\frac{k^{\prime }}{3}$

$\Rightarrow k^{\prime }=3k$

The equivalent spring constant of parallel combination of these spring is

$k^{\prime \prime }=k^{\prime }+k^{\prime }+k^{\prime }=3k^{\prime }=9k$

The time period of spring mass system of original spring $T=2\pi \sqrt{\frac{m}{k}}$, where, $m$=MAss attached to spring

The time period of parallel combination spring mass system,

$T^{\prime \prime }=2\pi \sqrt{\frac{m}{k^{\prime \prime }}}=2\pi \sqrt{\frac{m}{9k}}=\frac{1}{3}\left(2\pi \sqrt{\frac{m}{k}}\right)$

$T^{\prime \prime }=\frac{T}{3}$