The motion of a mass on a spring, with spring constant K is as shown… | JEE Main 2021 (22 Jul, Shift 2)

Q69JEE Main 2021 · 22 JUL · SHIFT 2Medium

Spring Systems

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 $\omega = \sqrt{\frac{K}{m}}$

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:

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Answer: (c)

$x(t) = C\cos(\omega t - \phi)$ , so $v(t) = -C\omega\sin(\omega t - \phi)$ .
At $t = 0$ : $x(0) = C\cos\phi$ and $v(0) = C\omega\sin\phi$ .
Square and add: $C = \sqrt{x(0)^2 + \frac{v(0)^2}{\omega^2}}$ . Divide: $\tan\phi = \frac{v(0)}{x(0)\omega}$ .
So the answer is (c).

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Spring Systems question from JEE Main 2021 (22 Jul, Shift 2)

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