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(B) The general equation for displacement in $SHM$ is given by $x(t) = a \cos(\omega t + \phi)$,where $a$ is the amplitude,$\omega$ is the angular fre

Vedclass · Accepted Answer

$x = a \cos(\frac{\pi}{2}t + \frac{\pi}{4})$

Vedclass · Answer

(B) The general equation for displacement in $SHM$ is given by $x(t) = a \cos(\omega t + \phi)$,where $a$ is the amplitude,$\omega$ is the angular frequency,and $\phi$ is the initial phase angle.Given the time period $T = 4 \ s$,the angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \ 	ext{rad/s}$.At $t = 0$,the particle is at an angle of $45^\circ$ with the $x$-axis,so the initial phase $\phi = 45^\circ = \frac{\pi}{4} \ 	ext{rad}$.Substituting these values into the general equation,we get $x(t) = a \cos(\frac{\pi}{2}t + \frac{\pi}{4})$.Thus,the correct option is $B$.

$A$ particle is executing simple harmonic motion $(SHM)$ with a time period $T = 4 \ s$. At $t = 0$,the particle is at a position where its angular position is $45^\circ$ with the $x$-axis. Find the displacement equation of the particle along the $x$-axis.

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