Name: Simple Harmonic Motion and Harmonic Oscillator
Uploaded: 2019-07-19
Duration: 9 min 10 s
Description: A harmonic oscillator is a system that, when displaced from its equilibrium position, experiences a restoring force F proportional to the displacement x.

Question

Derive expression for acceleration and time period in SHM.

Accepted Answer

Acceleration. The time rate of change of velocity is called acceleration.
The velocity of a particle executing S.H.M. at any instant ‘t’ having angular velocity ω and amplitude ‘r’ is given by:
$α = d v / dt = \frac{d}{d t} (r ω c o s ω t) = r ω \frac{d}{d t} (\cos ω t) α = r ω (- \sin ω t) ω = - ω^{2} x I n m a g n i t u d e, a = ω^{2} x . . . (i i) A t t h e m e a n p o s i t i o n, x = 0 a = 0 A t t h e e x t r e m e p o s i t i o n, x = r a = ω^{2} r = m a x i m u m$

Hence, the acceleration of a particle executing S.H.M. is maximum at the extreme position and zero at the mean position.

$T i m e p e r i o d : - F r o m e q u a t i o n a = ω^{2} x ω^{2} = \frac{a}{x} ω = \sqrt{\frac{a}{x}} \frac{2 π}{T} = \sqrt{\frac{a}{x}} T = 2 π \sqrt{\frac{x}{a}}$

also, $F = K x$ and $F = m a$

$∴ m a = K x$

$a = \frac{K x}{m}$

Therefore, time period, $T = 2 π \sqrt{\frac{x . m}{K x}} = 2 π \sqrt{\frac{m}{K}}$

Derive expression for acceleration and time period in SHM.

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