The velocity of a particle executing simple harmonic motion is $16\;cm/s$ at a distance of $8\;cm$ from the mean position and $8\;cm/s$ at a distance of $12\;cm$ from the mean position. Calculate the amplitude of the motion.

Given,

For $1^{st}$ Condition,

Velocity $v = 16\;cm/s$

Displacement $y = 8\;cm$

$v = \omega \sqrt{r^2 - y^2}$

$16 = \omega \sqrt{r^2 - 64}$ .......... (i)

Amplitude of the motion $r = \;?$

For $2^{nd}$ Condition,

Velocity $v = 8\;cm/s$

Displacement $y = 12\;cm$

$v = \omega \sqrt{r^2 - y^2}$

$8 = \omega \sqrt{r^2 - 144}$ .......... (ii)

From equation (i) and (ii), we get

$ \frac{16}{8} = \frac {\sqrt{r^2 - y^2}}{\sqrt{r^2 - 64}}$

or, $4 = $ $\frac{r^2\; - \;64}{r^2 \; - \;144}$

or, $4\;r^2 - 144 \;*\;4 = r^2 \;-\;64$

or, $3\;r^2 = 512$

or, $r^2 = 170.66\;cm$

or, $r = 13.06\;cm$

∴The amplitude of the motion is $13.06\;cm$

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