Question

A spring is cut into 5 equal parts & all of
them joined parallel to same mass m
. Find ratio of old & new time period​

Answer 1

Answer:

Given:

Spring cut into 5 equal parts and all of them joined parallel to same mass m.

To find:

Ratio of old and new time period.

Calculation:

Old time period

$t1 = 2\pi \sqrt{ \dfrac{m}{k} }$

Now since the spring has been cut, let the new spring constant be k2

$l1 \times k1 = l2 \times k2$

$= > l \times k = \frac{l}{5} \times k2$

$= > k2 = 5k$

Now, all the springs are attached in parallel.

$eq. \: spring \: constant = 5 \times 5$

$= > eq. \: spring \: constant =25$

Hence new time period

$t2 = 2\pi \sqrt{ \dfrac{m}{(25k)} }$

$= > t2 = \dfrac{1}{5} (2\pi \sqrt{ \dfrac{m}{k} } )$

Now, ratio:

$\dfrac{t1}{t2} = \dfrac{1}{ (\frac{1}{5}) }$

$= > ratio = 5: \: 1$

So final answer is

$\boxed{ \red{ratio \: = 5 : 1}}$

Answer 2

Answer:

Let initial length be L and spring constant be k.

And final length be L/5 and spring constant be k2

Whenever we cut spring in a particular ratio, we can use the following equality :

L × k = L/5 × k2

=> k2 = 5k

Now all the smaller springs are arranged parallel to each other .

Hence spring constant (net) = 5+5+5+5+5 = 25

Final time period = 2π√(m/25k)

Initial time period = 2π√(m/k)

Ratio of initial to final is

= 1 : (1/5)

= 5 : 1

So the answer is

$\boxed{ \huge{ \red{ratio = 5 : 1}}}$