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:

  1. 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, $T=\frac{2}{3} s$



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