First slide

Simple harmonic motion

Question

When a particle is performing linear SHM its KE is two times its PE at a position ‘A’ and its PE is two times its KE at another position ‘B’. Find ratio of ${KE}_{A}$ to ${KE}_{B}$

Moderate

A

$\frac{3}{2}$
B

$\frac{2}{1}$
C

$\frac{3}{4}$
D

$\frac{5}{2}$

Solution

$a t p o s i t i o n A, K E = 2 P E \Rightarrow P E = \frac{K E}{2}, T E = K E + P E \Rightarrow T E = K E + \frac{K E}{2} \Rightarrow T E_{A} = \frac{3 K E_{A}}{2} a t p o s i t i o n B, P E = 2 K E \Rightarrow T E = K E + P E \Rightarrow T E = K E + 2 K E \Rightarrow T E_{B} = 3 K E_{B} B y p r i n c i p l e o f c o n s e r v a t i o n o f e n e r g y T E_{A} = T E_{B} O n e q u a t i n g t h e t o t a l e n e r g y e q u a t i o n s, \frac{K E_{A}}{K E_{B}} = \frac{2}{1}$

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