Question

please answer this que​

Answer 1

The displacement is given by,

$\sf{\longrightarrow x=\left[5\sin(\pi t)+12\sin\left(\pi t+\dfrac{\pi}{2}\right)\right]\ cm}$

We see that,

$\sf{\longrightarrow\sin\left(\pi t+\dfrac{\pi}{2}\right)=\sin\left(\dfrac{\pi}{2}+\pi t\right)}$

$\sf{\longrightarrow\sin\left(\pi t+\dfrac{\pi}{2}\right)=\cos\left(\pi t\right)}$

Then,

$\sf{\longrightarrow x=\left[5\sin(\pi t)+12\cos\left(\pi t\right)\right]\ cm}$

Divide both sides by $\sf{\sqrt{5^2+12^2}=13.}$

$\sf{\longrightarrow\dfrac{x}{13}=\left[\dfrac{5}{13}\sin(\pi t)+\dfrac{12}{13}\cos\left(\pi t\right)\right]\ cm}$

Let there exist an acute angle $\alpha$ such that $\sf{\sin\alpha=\dfrac{12}{13}}$ and $\sf{\cos\alpha=\dfrac{5}{13}.}$

• $\sf{\alpha=\sin^{-1}\left(\dfrac{12}{13}\right)=\cos^{-1}\left(\dfrac{5}{13}\right)=1.176\ rad}$

Then,

$\sf{\longrightarrow\dfrac{x}{13}=\left[\sin(\pi t)\cos\alpha+\cos\left(\pi t\right)\sin\alpha\right]\ cm}$

Since $\sf{\sin x\cos y+\cos x\sin y=\sin(x+y),}$

$\sf{\longrightarrow\dfrac{x}{13}=\sin(\pi t+\alpha)\ cm}$

$\sf{\longrightarrow x=13\sin\left(\pi t+\sin^{-1}\left(\dfrac{12}{13}\right)\right)\ cm}$

or,

$\sf{\longrightarrow x=13\sin(\pi t+1.176)\ cm}$

Then,

• Amplitude $\bf{=13\ cm.}$

• Period $\bf{=\dfrac{2\pi}{\pi}=2.}$

[Period of the function $\sf{f(x)=\sin(ax+b)}$ is $\sf{\dfrac{2\pi}{a}}$ where $\sf{a\in\mathbb{R}-\{0\}}$ and $\sf{b\in\mathbb{R}.}$]

• Initial phase of motion $\bf{=\sin^{-1}\left(\dfrac{12}{13}\right)=1.176\ rad.}$