First slide

Simple harmonic motion

Question

Two springs of spring constants 1500 N/m and 3000 N/m respectively are stretched with the same force. They will have potential energy in the ratio

Moderate

A

4:1
B

1:4
C

2:1
D

1:2

Solution

$Potential energy PE= \frac{1}{2} {kx}^{2} R e s t o r i n g F o r c e o f s p r i n g F = - K x d i s p l a c e m e n t \Rightarrow x = \frac{F}{K} S u b s t i t u t e x v a l u e i n P E, P E = \frac{1}{2} K {(\frac{F}{K})}^{2} P E = \frac{1}{2} K \frac{F^{2}}{K^{2}} P E = \frac{1}{2} F^{2} /K \frac{P E_{1}}{P E_{2}} = \frac{K_{2}}{K_{1}} = 3000 / 1500 P E_{1} : P E_{2} = 2 : 1$

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