First slide

Applications of SHM

Question

Two springs of force constant 3000 N/m and 6000 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

$P o t e n t i a l E n e r g y P E o f s i m p l e h a r m o n i c o s c i l l a t o r i s P E = \frac{1}{2} {kx}^{2}, K=Force constant, x=displacement w e k n o w t h a t r e s t o r i n g f o r c e o f a s p r i n g i s F = - k x, h e r e k i s f o r c e c o n s \tan t, x i s d i s p l a c e m e n t \Rightarrow x=-F/k, Substitute this in PE, P E = \frac{1}{2} k {(- \frac{F}{k})}^{2} \Rightarrow P E = \frac{1}{2} k \frac{F^{2}}{k^{2}} \Rightarrow P E = \frac{1}{2} \frac{F^{2}}{k} \Rightarrow P E α \frac{1}{k}, b y q u e s t i o n F o r c e i s s a m e, h e n c e t a k e n c o n s \tan t, s u b s t i t u t e f o r c e c o n s \tan t k_{1}, k_{2} v a l u e s \Rightarrow \frac{P E_{1}}{P E_{2}} = \frac{k_{2}}{k_{1}} = \frac{6000}{3000} = \frac{2}{1}$

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