First slide

Simple hormonic motion

Question

A particle executes simple harmonic motion with time period 4 second. If it starts its motion from extreme position, then the time interval after which its kinetic energy will be equal to its potential energy for the first time is

Moderate

A

1.5 S
B

0.75 S
C

1.25 S
D

0.5 S

Solution

Equation of SHM is $x = Acosωt$
$∴$ Kinetic energy, $K = \frac{1}{2} {mV}^{2} = \frac{1}{2} {mA}^{2} ω^{2} \sin^{2} ωt$
Potential energy, $U = \frac{1}{2} {mω}^{2} x^{2} = \frac{1}{2} {mω}^{2} A^{2} \cos^{2} ωt$
When K = U, we get $\sin^{2} ωt = \cos^{2} ωt \Rightarrow tanωt = \pm 1$
$∴ ωt = \frac{π}{4}, \frac{3 π}{4}, 5 \frac{π}{4}, \frac{7 π}{4}, . ....... etc .$
$∴ \frac{2 π}{T} t_{\min} = \frac{π}{4} \Rightarrow t_{\min} = \frac{T}{8} = \frac{4}{8} S = 0 .5 S$

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