(B) Given: mass $m = 250\,g = 0.25\,kg$,force $F = -25x$,and maximum speed $v_{max} = 4\,m/s$.According to Newton's second law,$F = ma$,so $ma = -25x$

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(B) Given: mass $m = 250\,g = 0.25\,kg$,force $F = -25x$,and maximum speed $v_{max} = 4\,m/s$.According to Newton's second law,$F = ma$,so $ma = -25x$,which gives $a = -\frac{25}{0.25}x = -100x$.Comparing this with the standard $SHM$ equation $a = -\omega^2 x$,we get $\omega^2 = 100$,so $\omega = 10\,rad/s$.The maximum speed in $SHM$ is given by $v_{max} = \omega A$.Substituting the values: $4 = 10 \times A$.Therefore,$A = 0.4\,m$.Converting to centimeters: $A = 0.4 \times 100 = 40\,cm$.

Class 11 Physics Oscillations Velocity of Simple Harmonic Motion

Medium

$A$ particle of mass $250\,g$ executes a simple harmonic motion under a periodic force $F = (-25\,x)\,N$. The particle attains a maximum speed of $4\,m/s$ during its oscillation. The amplitude of the motion is $...........\,cm$.

JEE MAIN 2023 All JEE MAIN

A
$20$
B
$40$
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$60$
D
$80$

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Velocity of Simple Harmonic Motion

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$A$ particle of mass $250\,g$ executes a simple harmonic motion under a periodic force $F = (-25\,x)\,N$. The particle attains a maximum speed of $4\,m/s$ during its oscillation. The amplitude of the motion is $...........\,cm$.

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Obtain the instantaneous velocity of a particle executing $SHM$.

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$A$ particle performing $SHM$ has a time period of $\frac{2 \pi}{\sqrt{3}} \,s$ and a path length of $4 \,cm$. The displacement from the mean position at which the magnitude of acceleration is equal to the magnitude of velocity is (in $\,cm$)

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