First slide

Simple harmonic motion

Question

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then its periodic time (in seconds) is

Moderate

A

$\frac{4π}{3}$
B

$\frac{8π}{3}$
C

$\frac{7π}{3}$
D

$\frac{3π}{8}$

Solution

Given A=5 cms

At this time (when x=4 cm) velocity and acceleration have same magnitude
$|V| = |a|$ ,
$|ω \sqrt{A^{2} {-x}^{2}}| = |- ω^{2} x|$
cancel angular velocity w on both sides
$\sqrt{A^{2} {-x}^{2}} = ω^{2} x$
$\sqrt{5^{2} - 4^{2}} = ω (4)$
$3 = ω (4)$
$ω= \frac{3}{4} rad/s$
So time period time period is
$T= \frac{2π}{ω} s u b s t i t u t e ω = \frac{3}{4}, v a l u e T = \frac{2 π}{\frac{3}{4}} = \frac{8π}{3} sec$ .

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