First slide

Applications of SHM

Question

Two springs of force constant 1000 N/m & 2000 N/m are stretched by same force. The ratio of their respective potential energies is

Moderate

A

2:1
B

1:2
C

4:1
D

1:4

Solution

$p o t e n t i a l e n e r g y P E = \frac{1}{2} K X^{2}; K = f o r c e c o n s \tan t; x = d i s p l a c e m e n t r e s t o r i n g f o r c e = F = - K X \Rightarrow \frac{F}{K} = - X; s u b s t i t u t e X v a l u e i n P E e q u a t i o n, P E = \frac{1}{2} K \frac{F^{2}}{K^{2}} = \frac{1}{2} \frac{F^{2}}{K} f o r t w o s p r i n g s, P E_{1} = \frac{1}{2} \frac{F^{2}}{K_{1}} P E_{2} = \frac{1}{2} \frac{F^{2}}{K_{2}} d i v i d e t h e a b o v e t w o e q u a t i o n s, o n e b y t h e o t h e r \frac{P E_{1}}{P E_{2}} = \frac{K_{2}}{K_{1}}; s u b s t i t u t e t h e g i v e n v a l e s o f K_{1}, K_{2} \frac{P E_{1}}{P E_{2}} = = \frac{2000}{1000} = \frac{2}{1}$

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