First slide

Simple harmonic motion

Question

Total energy of particle performing SHM depends on

Moderate

A

amplitude and time period
B

amplitude, time period, displacement
C

amplitude displacement
D

time period, displacement

Solution

$t o t a l e n e r g y = \frac{1}{2} k A^{2} k = m ω^{2} = p r o p o g a t i o n c o n s \tan t A = a m p l i t u d e ω = \frac{2 π}{T} T = t i m e p e r i o d t o t a l e n e r g y = \frac{1}{2} m ω A^{2} T E = \frac{1}{2} m {[\frac{2 π}{T}]}^{2} A^{2} T E d e p e n d s o n m a s s o f o s c i l l a t o r, T i m e p e r i o d, A m p l i t u d e o f o s c i l l a t o r$

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