The time period of a mass suspended from a spring is $T$. If the spring is cut into three equal parts and connected in parallel. The same mass is suspended from these parallel springs, then the new time period of the mass will be

• A

  $\frac{T}{4}$
• B

  $T$
• Correct answer
  C

  $\frac{T}{3}$
• D

  $3 T$

Solution

As, spring constant, $(k) \propto \frac{1}{\text { length of spring }(l)}$ $\because$ The spring cut into three equal parts $\therefore k^{\prime}=3 k$ (For each spring) We know that, time period of a body connected with spring. $T=2 \pi \sqrt{\frac{m}{k}}$ ...(i) The parallel arrangement of the spring is given as
$\begin{aligned} & T^{\prime}=\begin{array}{l}\text { Time period of springs }\left(k^{\prime}=3 k\right) \text { connected in } \\ \text { parallel }\end{array} \\ & \qquad \\ & \qquad=2 \pi \sqrt{\frac{m}{3 k^{\prime}}}=2 \pi \sqrt{\frac{m}{3 \times 3 k}} \\ & \\ & \\ & =\frac{1}{3}\left(2 \pi \sqrt{\frac{m}{k}}\right)=\frac{T}{3} \quad \text { [From Eq. (i)] }\end{aligned}$

Hence, the correct answer is option: 'C'.