Question 11: If \mathrm{x}=5 \sin \left(\pi \mathrm{t}+\frac{\pi}{3}\right) \mathrm{m} represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion respectively, are :
(1) 5 \mathrm{~cm}, 2 \mathrm{~s}
(2) 5 \mathrm{~m}, 2 \mathrm{~s}
(3) 5 \mathrm{~cm}, 1 \mathrm{~s}
(4) 5 \mathrm{~m}, 1 \mathrm{~s}
Answer: Option (2)
Explanation:
The general equation of simple harmonic motion is
x=A\sin(\omega t+\phi),
where A is the amplitude, \omega is the angular frequency,
and \phi is the phase constant.
Comparing with the given equation x=5\sin\left(\pi t+\frac{\pi}{3}\right),
we get, A=5\ \mathrm{m}.
The angular frequency is \omega=\pi\ \mathrm{rad\,s^{-1}}.
The time period of SHM is given by
T=\frac{2\pi}{\omega}.
Substituting the value of \omega,
T=\frac{2\pi}{\pi}=2\ \mathrm{s}.
Hence, the amplitude is 5\ \mathrm{m}
and the time period is 2\ \mathrm{s}.
Therefore, the correct answer is option (2).