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Simple Harmonic Motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsSimple Harmonic Motion5 Marks
Question
A particle executes simple harmonic motion with an amplitude of 10cm and time period 6s. At t = 0 it is at position x = 5cm going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t = 4s.
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Answer

Given, $\text{r}=10\text{cm}.$ At $\text{t}=0,\ \text{x}=5\text{cm}$$\text{T}=6\sec.$
So, $\text{W}=\frac{2\pi}{\text{T}}=\frac{2\pi}{6}=\frac{\pi}{3}\sec^{-1}$ At, $\text{t}=0,\ \text{x}=5\text{cm}$ So, $\text{5}=10\sin(\text{w}\times0+\phi)=10\sin\phi$ $[\text{y}=\text{r}\sin\text{wt}]$$\sin\phi=\frac{1}{2}\Rightarrow\phi=\frac{\pi}{6}$
$\therefore$ Equation of displacement $\text{x}=(10\text{cm})\sin\Big(\frac{\pi}{3}\Big)$
At $\text{t}=4$ secound$\text{x}=10\sin\Big[\frac{\pi}{3}\times4+\frac{\pi}{6}\Big]=10\sin\Big[\frac{8\pi+\pi}{6}\Big]$
$=10\sin\Big(\frac{3\pi}{2}\Big)=10\sin\Big(\pi+\frac{\pi}{2}\Big)$
$=-10\sin\Big(\frac{\pi}{2}\Big)=-10$
Acceleration $\text{a}=-\text{w}^2\text{x}=-\Big(\frac{\pi^2}{9}\Big)\times(-10)$$=10.9\approx0.11\text{cm}/\sec$

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