Question

A spring block system in horizontal oscillation has a time period T. Now the spring is cut into 4 equal parts and the block is re-connected with one of the parts. The new time period of vertical oscillation will be

(1) $T/\sqrt{2}$
(2) $2T$
(3) $T/2$
(4) $T/2\sqrt{2}$

Answer 1

Option (3) The new time period of oscillation will be T/2

Let the spring constant of the initial spring is K.

Since, It is cut into four equal parts, All of them have a same spring constant.

Let the new spring constant be k

Consider they are arranged in series combination.

We can write,

⇒1/k + 1/k + 1/k + 1/k = 1/K

⇒ 4/k = 1/K

⇒ k = 4K

So we have now a new spring constant for each of the piece, and it is 4K

Time period of a the spring block system is given by,

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

The block isn't changed. So m remains same.

$T \propto \frac{1}{ \sqrt{k} }$

Given, Initial Time period = T

Let the new time period be t

$\frac{T}{t} = \sqrt{ \dfrac{k}{K}} \\ \\ \frac{T}{t} = \sqrt{ \frac{4K}{K} } \\ \\ \frac{T}{t} = \sqrt{4} \\ \\ t = \frac{T}{2}$

Therefore, The new time period is T/2.

Answer 2

Answer:

(c) T/2

Explanation:

Time period for spring is given as:

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

where, k is the spring constant of the spring.

When the spring is cut into 4 parts, the equivalent spring constant of each part becomes 4k.

Check out the attachment for the detailed solution.

Thanks!