First slide

Simple hormonic motion

Question

A particle is executing SHM along a straight line. Its velocities at distances $x_{1} and x_{2}$ from the mean position are $V_{1} and V_{2}$ respectively. Its time period is

Moderate

A

$2 π \sqrt{\frac{{x^{2}}_{1} + {x^{2}}_{2}}{{V^{2}}_{1} + {V^{2}}_{2}}}$
B

$2 π \sqrt{\frac{{x^{2}}_{2} - {x^{2}}_{1}}{{V^{2}}_{1} - {V^{2}}_{2}}}$
C

$2 π \sqrt{\frac{{V^{2}}_{1} + {V^{2}}_{2}}{{x^{2}}_{1} + {x^{2}}_{2}}}$
D

$2 π \sqrt{\frac{{V^{2}}_{1} - {V^{2}}_{2}}{{x^{2}}_{1} - {x^{2}}_{2}}}$

Solution

As we know, for particle undergoing SHM,
$V = ω \sqrt{A^{2} - X^{2}}$
${V^{2}}_{1} = ω^{2} (A^{2} - {x^{2}}_{1}) and {V^{2}}_{2} = ω^{2} (A^{2} - {x^{2}}_{2})$
On subtracting the relations
${V^{2}}_{1} - {V^{2}}_{2} = ω^{2} ({x^{2}}_{2} - {x^{2}}_{1})$
$ω = \sqrt{\frac{{V^{2}}_{1} - {V^{2}}_{2}}{{x^{2}}_{2} - {x^{2}}_{1}}}$
$T = 2 π \sqrt{\frac{{x^{2}}_{2} - {x^{2}}_{1}}{{V^{2}}_{1} - {V^{2}}_{2}}}$

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