(B) Given,initial position $x = 1 \, cm$,initial velocity $v = \pi \, cm/s$,and angular frequency $\omega = \pi \, s^{-1}$.Using the relation between

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(B) Given,initial position $x = 1 \, cm$,initial velocity $v = \pi \, cm/s$,and angular frequency $\omega = \pi \, s^{-1}$.Using the relation between velocity and displacement in $SHM$:$v = \omega \sqrt{A^2 - x^2}$Substituting the given values:$\pi = \pi \sqrt{A^2 - (1)^2}$Dividing both sides by $\pi$:$1 = \sqrt{A^2 - 1}$Squaring both sides:$1 = A^2 - 1$$A^2 = 2$$A = \sqrt{2} \, cm$.Thus,the amplitude is $\sqrt{2} \, cm$.

Class 11 Physics Oscillations Velocity of Simple Harmonic Motion

Easy

$A$ particle in $SHM$ is described by the displacement equation $x(t) = A\cos(\omega t + \theta)$. If the initial $(t = 0)$ position of the particle is $1 \, cm$ and its initial velocity is $\pi \, cm/s$,what is its amplitude? The angular frequency of the particle is $\pi \, s^{-1}$.

A
$1 \, cm$
B
$\sqrt{2} \, cm$
C
$2 \, cm$
D
$2.5 \, cm$

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Oscillations

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Velocity of Simple Harmonic Motion

The maximum velocity of a body undergoing $S.H.M.$ is $0.2 \ m/s$ and its acceleration at $0.1 \ m$ from the mean position is $0.4 \ m/s^2$. The amplitude of the $S.H.M.$ is .... $m$

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$A$ particle executes linear simple harmonic motion with an amplitude of $3\,cm$. When the particle is at $2\,cm$ from the mean position,the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:

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The velocity of a particle in simple harmonic motion at displacement $y$ from the mean position is

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The amplitude of an oscillating particle is $A$. When the velocity of the particle is one-third of its maximum velocity,determine the position of the particle.

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The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of $1.0 \; m$. If the piston moves with simple harmonic motion with an angular frequency of $200 \; rad/min$,what is its maximum speed in $m/min$?

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$A$ particle in $SHM$ is described by the displacement equation $x(t) = A\cos(\omega t + \theta)$. If the initial $(t = 0)$ position of the particle is $1 \, cm$ and its initial velocity is $\pi \, cm/s$,what is its amplitude? The angular frequency of the particle is $\pi \, s^{-1}$.

Similar Questions

The maximum velocity of a body undergoing $S.H.M.$ is $0.2 \ m/s$ and its acceleration at $0.1 \ m$ from the mean position is $0.4 \ m/s^2$. The amplitude of the $S.H.M.$ is .... $m$

$A$ particle executes linear simple harmonic motion with an amplitude of $3\,cm$. When the particle is at $2\,cm$ from the mean position,the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:

The velocity of a particle in simple harmonic motion at displacement $y$ from the mean position is

The amplitude of an oscillating particle is $A$. When the velocity of the particle is one-third of its maximum velocity,determine the position of the particle.

The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of $1.0 \; m$. If the piston moves with simple harmonic motion with an angular frequency of $200 \; rad/min$,what is its maximum speed in $m/min$?

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