A body executing S.H.M. along a straight line | vedclass

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(D) The velocity $v$ of a particle in $S.H.M.$ at a displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $\omega$ is the angular frequency

Vedclass · Accepted Answer

$1 \, rad/s$ and $5 \, m$

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(D) The velocity $v$ of a particle in $S.H.M.$ at a displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $\omega$ is the angular frequency and $A$ is the amplitude.Given:For $x_1 = 4 \, m$,$v_1 = 3 \, m/s$. Thus,$3 = \omega \sqrt{A^2 - 4^2} \implies 9 = \omega^2 (A^2 - 16) \quad (i)$For $x_2 = 3 \, m$,$v_2 = 4 \, m/s$. Thus,$4 = \omega \sqrt{A^2 - 3^2} \implies 16 = \omega^2 (A^2 - 9) \quad (ii)$Dividing equation $(i)$ by $(ii)$:$\frac{9}{16} = \frac{\omega^2 (A^2 - 16)}{\omega^2 (A^2 - 9)} = \frac{A^2 - 16}{A^2 - 9}$$9(A^2 - 9) = 16(A^2 - 16)$$9A^2 - 81 = 16A^2 - 256$$7A^2 = 175 \implies A^2 = 25 \implies A = 5 \, m$Substituting $A^2 = 25$ into equation $(i)$:$9 = \omega^2 (25 - 16) = 9\omega^2$$\omega^2 = 1 \implies \omega = 1 \, rad/s$Thus,the angular frequency is $1 \, rad/s$ and the amplitude is $5 \, m$.

$A$ body executing $S.H.M.$ along a straight line has a velocity of $3 \, m/s$ when it is at a distance of $4 \, m$ from its mean position and $4 \, m/s$ when it is at a distance of $3 \, m$ from its mean position. Its angular frequency and amplitude are:

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