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Question

(A) The displacement equation for $SHM$ is $x(t) = A \sin(\omega(t - t_0))$,where $t_0$ is the time at which the particle is at the equilibrium positi

Vedclass · Accepted Answer

$\frac{3}{2\pi} \, m$

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(A) The displacement equation for $SHM$ is $x(t) = A \sin(\omega(t - t_0))$,where $t_0$ is the time at which the particle is at the equilibrium position.Given $t_0 = 1 \, s$,so $x(t) = A \sin(\omega(t - 1))$.The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3} \, rad/s$.The velocity is $v(t) = \frac{dx}{dt} = A\omega \cos(\omega(t - 1))$.At $t = 2 \, s$,$v = 0.25 \, m/s = \frac{1}{4} \, m/s$.Substituting the values: $\frac{1}{4} = A \left(\frac{\pi}{3}\right) \cos\left(\frac{\pi}{3}(2 - 1)\right)$.$\frac{1}{4} = A \left(\frac{\pi}{3}\right) \cos\left(\frac{\pi}{3}\right)$.Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we have $\frac{1}{4} = A \left(\frac{\pi}{3}\right) \left(\frac{1}{2}\right)$.$\frac{1}{4} = A \left(\frac{\pi}{6}\right)$.$A = \frac{6}{4\pi} = \frac{3}{2\pi} \, m$.

$A$ particle performing $SHM$ is found at its equilibrium position at $t = 1 \, s$. It is found to have a speed of $0.25 \, m/s$ at $t = 2 \, s$. If the period of oscillation is $6 \, s$,calculate the amplitude of oscillation.

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