(B) Given: Amplitude $A = 4 \ cm$,Displacement $x = 3 \ cm$.Velocity in $S.H.M.$ is given by $v = \omega \sqrt{A^2 - x^2}$.Acceleration in $S.H.M.$ is

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(B) Given: Amplitude $A = 4 \ cm$,Displacement $x = 3 \ cm$.Velocity in $S.H.M.$ is given by $v = \omega \sqrt{A^2 - x^2}$.Acceleration in $S.H.M.$ is given by $a = \omega^2 x$.According to the problem,the magnitude of velocity equals the magnitude of acceleration: $|v| = |a|$.$\omega \sqrt{A^2 - x^2} = \omega^2 x$.$\sqrt{A^2 - x^2} = \omega x$.Substitute the values: $\sqrt{4^2 - 3^2} = \omega (3)$.$\sqrt{16 - 9} = 3 \omega$.$\sqrt{7} = 3 \omega$.$\omega = \frac{\sqrt{7}}{3} \ rad/s$.The time period $T$ is given by $T = \frac{2 \pi}{\omega}$.$T = \frac{2 \pi}{\frac{\sqrt{7}}{3}} = \frac{6 \pi}{\sqrt{7}} \ s$.

Class 11 Physics Oscillations Method to determine Time Period and Frequency for diffrent type of Object of SHM

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$A$ particle executes linear $S.H.M.$ with amplitude $4 \ cm$. The magnitude of velocity and acceleration is equal when it is at $3 \ cm$ from the mean position. The time period of oscillation is:

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A
$\frac{\pi}{\sqrt{7}} \ s$
B
$\frac{6 \pi}{\sqrt{7}} \ s$
C
$\frac{3 \pi}{\sqrt{7}} \ s$
D
$\frac{5 \pi}{\sqrt{7}} \ s$

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Method to determine Time Period and Frequency for diffrent type of Object of SHM

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$A$ body executes simple harmonic motion under the action of a force $F_1$ with a time period $(4/5) \ s$. If the force is changed to $F_2$,it executes $SHM$ with a time period $(3/5) \ s$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body,its time period (in $s$) is:

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$A$ $U$-tube of a uniform bore has its arms vertical. The total length of the liquid in the two arms of the $U$-tube is $L$. The time period $T$ of the oscillation of the liquid column when it is displaced by $y$ is ($g$ = acceleration due to gravity).

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The potential energy of a particle of mass $1\,kg$ in motion along the $x-$axis is given by $U = 4(1 - \cos 2x)\,J,$ where $x$ is in $meters$. The period of small oscillations (in $sec$) is

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The height of a liquid column in a $U$-tube is $0.3 \ m$. If the liquid in one of the limbs is depressed and then released,the time period of the liquid column will be $....... \ s$.

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$A$ rectangular block of mass '$m$' and cross-sectional area '$A$' floats on a liquid of density '$\rho$'. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency ($g$ = acceleration due to gravity).

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$A$ particle executes linear $S.H.M.$ with amplitude $4 \ cm$. The magnitude of velocity and acceleration is equal when it is at $3 \ cm$ from the mean position. The time period of oscillation is:

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$A$ $U$-tube of a uniform bore has its arms vertical. The total length of the liquid in the two arms of the $U$-tube is $L$. The time period $T$ of the oscillation of the liquid column when it is displaced by $y$ is ($g$ = acceleration due to gravity).

The potential energy of a particle of mass $1\,kg$ in motion along the $x-$axis is given by $U = 4(1 - \cos 2x)\,J,$ where $x$ is in $meters$. The period of small oscillations (in $sec$) is

The height of a liquid column in a $U$-tube is $0.3 \ m$. If the liquid in one of the limbs is depressed and then released,the time period of the liquid column will be $....... \ s$.

$A$ rectangular block of mass '$m$' and cross-sectional area '$A$' floats on a liquid of density '$\rho$'. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency ($g$ = acceleration due to gravity).

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