Question

The motion of a mass on a spring, with spring constant \(K\) is as shown in figure.

The equation of motion is given by
\(x(t) = A\sin\omega t + B\cos\omega t\) with \(\omega = \sqrt{\dfrac{K}{m}}\)

Suppose that at time \(t = 0\), the position of mass is \(x(0)\) and velocity \(v(0)\), then its displacement can also be represented as \(x(t) = C\cos(\omega t - \phi)\), where \(C\) and \(\phi\) are :

• A. \(C = \sqrt{\dfrac{2v(0)^2}{\omega^2} + x(0)^2},\ \phi = \tan^{-1}\left(\dfrac{x(0)\omega}{2v(0)}\right)\)
• B. \(C = \sqrt{\dfrac{v(0)^2}{\omega^2} + x(0)^2},\ \phi = \tan^{-1}\left(\dfrac{x(0)\omega}{v(0)}\right)\)
• C. \(C = \sqrt{\dfrac{v(0)^2}{\omega^2} + x(0)^2},\ \phi = \tan^{-1}\left(\dfrac{v(0)}{x(0)\omega}\right)\)
• D. \(C = \sqrt{\dfrac{2v(0)^2}{\omega^2} + x(0)^2},\ \phi = \tan^{-1}\left(\dfrac{v(0)}{x(0)\omega}\right)\)

Hint:

Find \(A\) and \(B\) from \(x(0)\) and \(v(0)\), then match \(A\sin\omega t + B\cos\omega t\) with \(C\cos\phi\cos\omega t + C\sin\phi\sin\omega t\).

Answer 1

The Correct Option is C

\(x(0) = B\). Differentiating, \(v(t) = A\omega\cos\omega t - B\omega\sin\omega t\), so \(v(0) = A\omega\) and \(A = v(0)/\omega\).
Expand \(C\cos(\omega t - \phi) = C\cos\phi\cos\omega t + C\sin\phi\sin\omega t\).
Comparing: \(C\cos\phi = B = x(0)\) and \(C\sin\phi = A = \dfrac{v(0)}{\omega}\).
Squaring and adding: \(C = \sqrt{x(0)^2 + \dfrac{v(0)^2}{\omega^2}}\).
Dividing: \(\tan\phi = \dfrac{v(0)}{\omega\,x(0)}\), so \(\phi = \tan^{-1}\left(\dfrac{v(0)}{x(0)\omega}\right)\).
Answer: (3) \(C = \sqrt{\dfrac{v(0)^2}{\omega^2} + x(0)^2},\ \phi = \tan^{-1}\left(\dfrac{v(0)}{x(0)\omega}\right)\)