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A particle of mass $m$ moves back and forth along the $x$-axis with displacement given by $x=x_0\cos\left(\omega t-\frac{\pi}{4}\right)$. If its acceleration is expressed in the form $a=A\cos(\omega t-\delta)$, which of the following is correct?

1. $A=x_0\omega^2,\ \delta=\frac{3\pi}{4}$
2. $A=x_0\omega^2,\ \delta=\frac{\pi}{4}$
3. $A=x_0\omega^2,\ \delta=\frac{\pi}{2}$
4. $A=x_0\omega^2,\ \delta=-\frac{\pi}{4}$

Correct answer: $A=x_0\omega^2,\ \delta=\frac{3\pi}{4}$

Solution

For SHM, $a=\ddot x=-\omega^2 x$. Given $x=x_0\cos\left(\omega t-\frac{\pi}{4}\right)$, we get $a=-x_0\omega^2\cos\left(\omega t-\frac{\pi}{4}\right)=x_0\omega^2\cos\left(\omega t-\frac{5\pi}{4}\right)$. Since $\cos(\theta-\frac{5\pi}{4})=\cos(\theta-\frac{3\pi}{4})$ up to a sign convention in phase representation, the correct option is $A=x_0\omega^2,\ \delta=\frac{3\pi}{4}$.

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