First slide

Applications of SHM

Question

Two springs of force constant 2000 N/m and 4000 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

$P E o f m a s s a t t a c h e d t o a s p r i n g i s P E = \frac{1}{2} {kx}^{2}, where k= force constant, x= displacement r e s t o r i n g f o r c e o f a s p r i n g i s F = - k x, \Rightarrow x=- \frac{F}{k},substitute in PE equation P E = \frac{1}{2} k {(\frac{F}{k})}^{2} \Rightarrow P E = \frac{1}{2} k \frac{F^{2}}{k^{2}} \Rightarrow P E = \frac{1}{2} \frac{F^{2}}{k} \Rightarrow P E α \frac{1}{k} \sin c e s t r e t c h e d b y s a m e f o r c e \Rightarrow \frac{P E_{1}}{{PE}_{2}} = \frac{k_{2}}{k_{1}} = \frac{4000}{2000} = \frac{2}{1} o b t a i n e d a f t e r s u b s t i t u t i n g t h e g i v e n v a l u e s o f f o r c e c o n s \tan t k_{1} {,k}_{2}$

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