First slide

Simple harmonic motion

Question

A particle executes SHM of time period T. The time taken by the particle to move from mean position to the position where its kinetic energy is equal to potential energy is

Moderate

A

$\frac{T}{3}$
B

$\frac{T}{4}$
C

$\frac{T}{6}$
D

$\frac{T}{8}$

Solution

$\begin{array}{l} x = Asin ω t \\ v = ω \sqrt{A^{2} - x^{2}} \\ So, KE = \frac{1}{2} {mv}^{2} = \frac{1}{2} m ω^{2} A^{2} - \frac{1}{2} m ω^{2} x^{2} \\ And PE = \frac{1}{2} {Kx}^{2} = \frac{1}{2} m ω^{2} x^{2} \\ PE = KE \Rightarrow \frac{1}{2} m ω^{2} A^{2} - \frac{1}{2} m ω^{2} x^{2} = \frac{1}{2} m ω^{2} x^{2} \\ \Rightarrow x = \pm \frac{A}{\sqrt{2}} \end{array}$
$\begin{array}{l} Let us consider motion from x = 0 to x = \frac{A}{\sqrt{2}} \\ \frac{A}{\sqrt{2}} = Asin ω t \\ ω t = \frac{π}{4}, \frac{3 π}{4}, \frac{9 π}{4}, \frac{11 π}{4}, \dots \\ \Rightarrow t = \frac{T}{8}, \frac{3 T}{8}, \frac{9 T}{8}, \frac{11 T}{8}, . . \end{array}$

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