(A) For a particle in simple harmonic motion,the maximum acceleration is given by $a_{\max} = \omega^2 A = \alpha$ $...(1)$The maximum velocity is giv

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(A) For a particle in simple harmonic motion,the maximum acceleration is given by $a_{\max} = \omega^2 A = \alpha$ $...(1)$The maximum velocity is given by $v_{\max} = \omega A = \beta$ $...(2)$Dividing equation $(1)$ by equation $(2)$,we get:$\frac{\omega^2 A}{\omega A} = \frac{\alpha}{\beta}$$\omega = \frac{\alpha}{\beta}$Since angular frequency $\omega = 2\pi f$,where $f$ is the frequency of vibration:$2\pi f = \frac{\alpha}{\beta}$$f = \frac{\alpha}{2\pi\beta}$

Class 11 Physics Oscillations Acceleration and Force of Simple Harmonic Motion

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$A$ particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$. Then its frequency of vibration will be

A
$\frac{\alpha}{2\pi\beta}$
B
$\frac{\alpha^2}{\beta^2}$
C
$\frac{\beta}{\alpha}$
D
$\frac{\alpha}{\beta^2}$

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Acceleration and Force of Simple Harmonic Motion

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The $x-t$ graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at $t=2 \ s$ is:

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$A$ particle vibrating simple harmonically has an acceleration of $16 \ cm/s^2$ when it is at a distance of $4 \ cm$ from the mean position. Its time period is: (in $s$)

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$A$ particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$. Then its frequency of vibration will be

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$A$ piston is performing $S.H.M.$ in the vertical direction with a frequency of $0.5 \,Hz$. $A$ block of $10 \,kg$ is placed on the piston. The maximum amplitude of the system such that the block remains in contact with the piston is: (in $\,m$)

Acceleration of a particle,executing $SHM$,at its mean position is

The $x-t$ graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at $t=2 \ s$ is:

$A$ point mass oscillates along the $x$-axis according to the law $x=x_0 \cos(\omega t - \frac{\pi}{4})$. If the acceleration of the particle is written as $a=A \cos(\omega t + \delta)$,then:

$A$ particle vibrating simple harmonically has an acceleration of $16 \ cm/s^2$ when it is at a distance of $4 \ cm$ from the mean position. Its time period is: (in $s$)

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