First slide

Simple hormonic motion

Question

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Moderate

A

$\frac{\sqrt{5}}{2 π}$
B

$\frac{4 π}{\sqrt{5}}$
C

$\frac{2 π}{\sqrt{3}}$
D

$\frac{\sqrt{5}}{π}$

Solution

Given, A = 3 cm, x = 2 cm
The velocity of a particle in simple harmonic motion is given as
$v = ω \sqrt{A^{2} - x^{2}}$
$and magnitude of its acceleration is a = ω^{2} x$
$Given | v | = | a |$
$∴ ω \sqrt{A^{2} - x^{2}} = ω^{2} x$
$ω x = \sqrt{A^{2} - x^{2}} or ω^{2} x^{2} = A^{2} - x^{2}$
$ω^{2} = \frac{A^{2} - x^{2}}{x^{2}} = \frac{9 - 4}{4} = \frac{5}{4}$
$ω = \frac{\sqrt{5}}{2}$
$Time period, T = \frac{2 π}{ω} = 2 π \cdot \frac{2}{\sqrt{5}} = \frac{4 π}{\sqrt{5}} s$

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