A body performs simple harmonic motion according to the following… - Vidyadip

Question

A body performs simple harmonic motion according to the following equation :
$
x=6 \sin \left(3 \pi t+\frac{\pi}{3}\right)
$
Find out : (i) amplitude (ii) period (iii) initial art (iv) displacement, velocity and acceleration at time $t = 2$.

✓

Answer

Given equation :
$x=6 \sin \left(3 \pi t+\frac{\pi}{3}\right)$
The general equation of simple harmonic motion is as follows :
$x=A \sin (\omega t+\phi)$
Comparing it with the given equation
(i) Dimension $A =6 m$
(ii) $\omega=3 \pi$
$\begin{aligned}\text {But}\quad\omega & =\frac{2 \pi}{T} \\T & =\frac{2 \pi}{\omega}=\frac{2 \pi}{3 \pi}=\frac{2}{3} sec \\T & =0 \cdot 666 sec .\end{aligned}$
(iii) Initial phase $=\frac{\pi}{3}$ radian
(iv) $x=6 \sin \left(3 \pi t +\frac{\pi}{3}\right)$ by keeping $t=2$ in
$\begin{aligned}x & =6 \sin \left(3 \pi \times 2+\frac{\pi}{3}\right) \\& =6 \sin \frac{\pi}{3}=6 \times \frac{\sqrt{3}}{2}=3 \sqrt{3}\end{aligned}$
Displacement $=3 \sqrt{3} m$.
$A =6$ and $\omega=3 \pi$ putting these values in the equation of velocity,
$\begin{array}{l}v= 3 \pi\left[(6)^2-(3 \sqrt{3})^2\right]^{\frac{1}{2}} \\\quad\left(\because v=\omega \sqrt{A^2-x^2}\right) \\=3 \pi(36-27)^{\frac{1}{2}}=3 \pi \times 3=9 \pi\end{array}$
Acceleration $a=\omega^2 x=(3 \pi)^2 \times 3 \sqrt{3}$
$=9 \pi^2 \times 3 \sqrt{3}=27 \sqrt{3} \pi^2$
Amplitude $=6$, period $=0.666 sec$.
Initial phase $=\frac{\pi}{3}$, displacement $=3 \sqrt{3} m / s$
Velocity $=9 \pi m / s$ and acceleration
$=27 \sqrt{3} \pi^2 m / s^2$

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