First slide

Simple harmonic motion

Question

The mean kinetic energy of a particle executing SHM over one full oscillation, given mass of particle = m, amplitude = A, angular frequency = $ω$ ).

Moderate

A

$\frac{1}{4} m ω^{2} A^{2}$
B

$\frac{1}{2} m ω^{2} A^{2}$
C

$\frac{1}{2 \sqrt{2}} m ω^{2} A^{2}$
D

$\frac{1}{8} m ω^{2} A^{2}$

Solution

$\begin{array}{l} x = Asin ω t So, v = A ω \cos ω t \\ KE = \frac{1}{2} {mv}^{2} = \frac{1}{2} m (A ω \cos ω t)^{2} = \frac{1}{2} m ω^{2} A^{2} \cos^{2} ω t \end{array}$
Mean kinetic energy =
$\begin{array}{l} {KE}_{mean} = \frac{\int_{0}^{T} (KE) dt}{\int_{\int}^{T} dt} = \frac{\int_{0}^{T} (\frac{1}{2} m ω^{2} A^{2} \cos^{2} ω t) dt}{T} = \frac{\frac{1}{2} m ω^{2} A^{2} T}{T} \int_{0}^{T} \cos^{2} ω tdt \\ = \frac{\frac{1}{2} m ω^{2} A^{2}}{(T)} \int_{0}^{T} (\frac{1 + \cos 2 ω t}{2}) dt = \frac{\frac{1}{2} m ω^{2} A^{2}}{T} {[\frac{t}{2} + \frac{\sin 2 ω t}{4 ω}]}_{0}^{T} \\ = \frac{\frac{1}{2} m ω^{2} A^{2}}{T} [\frac{T}{2}] = \frac{1}{4} m ω^{2} A^{2} . \end{array}$

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