Simple Harmonic Motion 3 Question 2
2. A simple harmonic motion is represented by $y=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t) cm$. The amplitude and time period of the motion are
(a) $10 cm, \frac{3}{2} s$
(b) $5 cm, \frac{2}{3} s$
(c) $5 cm, \frac{3}{2} s$
(d) $10 cm, \frac{2}{3} s$
(2019 Main, 12 Jan II)
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Answer:
Correct Answer: 2. (d)
Solution:
- Equation for $SHM$ is given as
$$ \begin{aligned} y & =5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t) \\ & =5 \times 2 \frac{1}{2} \times \sin 3 \pi t+\frac{\sqrt{3}}{2} \cos 3 \pi t \\ & =5 \times 2 \cos \frac{\pi}{3} \cdot \sin 3 \pi t+\sin \frac{\pi}{3} \cdot 3 \pi t \\ & =5 \times 2 \sin 3 \pi t+\frac{\pi}{3} \\ & \quad[\text { using, } \sin (a+b)=\sin a \cos b+\cos a \sin b] \end{aligned} $$
$$ \text { or } \quad y=10 \sin 3 \pi t+\frac{\pi}{3} $$
Comparing this equation with the general equation of SHM, i.e.
$$ y=A \sin \frac{2 \pi t}{T}+\varphi $$
We get, amplitude, $A=10 cm$
and $3 \pi=\frac{2 \pi}{T}$
or Time period, $T=\frac{2}{3} s$
