If $x= 5sin (\pi t+\frac{\pi}{3})$m represents the motion of a partcile executing simple harmonic motion, the amplitude and time period of motion, respectively, are :
Answer & explanation
Correct answer: option 2
The correct answer is option (2) : 5 m, 2s
In the equation for simple harmonic motion (SHM), $x= 5 sin (\pi t+\frac{\pi}{3})m$, the general form $x= A sin (\omega t + \phi ) $ can be used to identify the parameters of SHM, where :
$A$ is the amplitude.
$\omega $ is the angular frequency.
$\phi $ is the phase constant.
$t$ is the time.
Comparing the given equation with the standard form :
The amplitude $A$ is $5\,m$, as that is the coefficient of sine in the equation.
The angular frequency $\omega $ is $\pi $ rad/s.
The angular frequency $\omega $ is related to the time period $T$ of the motion through the formula :
$\omega =\frac{2\pi }{T}$
Given that $\omega = \pi , $ we can substitute and solve for $T$ :
$\pi =\frac{2\pi}{T}⇒T=\frac{2\pi}{\pi }= 2 $ seconds .
Hence, the amplitude of the motion is 5 m and the time period is 2 s. Thus, the correct answer is :
Option 2 : 5m , 2s