A particle executing linear S.H.M. has velocities V 1 and V 2 at distance x 1 and x 2 respectively, from the mean position, its angular velocity is

Question

A particle executing linear S.H.M. has velocities V 1 ​ and V 2 ​ at distance x 1 ​ and x 2 ​ respectively, from the mean position, its angular velocity is

TestHub · Accepted Answer

The velocity of a particle executing linear S.H.M. at a distance $x$ from the mean position is given by $V = ω A^{2} - x^{2}$ , where $A$ is the amplitude and $ω$ is the angular velocity.

Squaring both sides, we get $V^{2} = ω^{2} (A^{2} - x^{2})$ .

For the two given positions, we can write:
$V_{1}^{2} = ω^{2} (A^{2} - x_{1}^{2})$
$V_{2}^{2} = ω^{2} (A^{2} - x_{2}^{2})$

Subtracting the first equation from the second, we get:
$V_{2}^{2} - V_{1}^{2} = ω^{2} (A^{2} - x_{2}^{2}) - ω^{2} (A^{2} - x_{1}^{2})$

$V_{2}^{2} - V_{1}^{2} = ω^{2} (x_{1}^{2} - x_{2}^{2})$

$ω^{2} = \frac{V _{2}^{2} - V _{1}^{2}}{x _{1}^{2} - x _{2}^{2}}$

$ω = \frac{V _{2}^{2} - V _{1}^{2}}{x _{1}^{2} - x _{2}^{2}}$

Answer: $\frac{V _{2}^{2} - V _{1}^{2}}{x _{1}^{2} - x _{2}^{2}}$

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