First slide

Simple hormonic motion

Question

A spring of force constant k is cut into lengths of ratio 1: 2: 3 . They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k". Then k' : k" is

Moderate

A

1 : 9
B

1 : 11
C

1 : 14
D

1 : 6

Solution

Let us assume, the length of spring be l. When we cut the spring into ratio of length 1: 2: 3, we get three springs of lengths $\frac{l}{6}, \frac{2 l}{6} and \frac{3 l}{6}$ with force constant,
$\begin{array}{l} ∴ k_{1} = \frac{k l}{l_{1}} = \frac{k l}{l / 6} = 6 k \\ k_{2} = \frac{k l}{l_{2}} = \frac{k l}{2 l / 6} = 3 k \\ k_{3} = \frac{k l}{l_{3}} = \frac{k l}{3 l / 6} = 2 k \end{array}$
When connected in series,
$\frac{1}{k^{'}} = \frac{1}{6 k} + \frac{1}{3 k} + \frac{1}{2 k} = \frac{1 + 2 + 3}{6 k} = \frac{1}{k}$
$∴ k^{'} = k$
When connected in parallel,
$k^{''} = 6 k + 3 k + 2 k = 11 k \frac{k^{'}}{k^{''}} = \frac{k}{11 k} = \frac{1}{11}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App