First slide

Simple hormonic motion

Question

A particle executes SHM on a straight line path. The amplitude of oscillation is 2 cm. When the displacement of the particle from the mean position is I cm, the numerical value of magnitude of acceleration is equal to the numerical value of magnitude of velocity. Then find out the frequency of SHM.

Easy

A

$\frac{\sqrt{3}}{2 π}$
B

$\frac{3}{2 π}$
C

$\frac{3}{\sqrt{2} π}$
D

$\frac{\sqrt{3}}{π}$

Solution

The speed of particle at a distance x = I from mean position is
$v = ω \sqrt{A^{2} - x^{2}} = ω \sqrt{2^{2} - 1^{2}} = \sqrt{3} s$ ...(i)
The magnitude of acceleration at x = I is
$a = ω^{2} x = ω^{2} - - - - - - - - (ii)$
from equation (i) and (ii)
$ω^{2} = \sqrt{3} ω \Rightarrow ω = \sqrt{3}$
or $f = \frac{ω}{2 π} = \frac{\sqrt{3}}{2 π}$

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