The angular frequency of a spring-block syste | vedclass

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(B) $1$. When the elevator is moving downwards at a constant speed $v_0$,the block is at rest relative to the elevator. When the elevator stops sudden

Vedclass · Accepted Answer

The initial phase of the block is $\pi$.

Vedclass · Answer

(B) $1$. When the elevator is moving downwards at a constant speed $v_0$,the block is at rest relative to the elevator. When the elevator stops suddenly,the block continues to move downwards with an initial velocity $v_0$ relative to the ground.$2$. The equilibrium position of the spring-block system remains unchanged because the gravitational force and spring force are balanced. The block now has an initial velocity $v_0$ at the equilibrium position.$3$. The amplitude $A$ is given by $v_{max} = A\omega_0$,so $A = \frac{v_0}{\omega_0}$. Thus,option $A$ is correct.$4$. Since the block starts from the equilibrium position moving in the positive (downward) direction,the equation of motion is $x(t) = A \sin(\omega_0 t) = \frac{v_0}{\omega_0} \sin(\omega_0 t)$. Thus,option $C$ is correct.$5$. The maximum speed in $SHM$ is $v_{max} = A\omega_0 = \frac{v_0}{\omega_0} \cdot \omega_0 = v_0$. Thus,option $D$ is correct.$6$. The initial phase $\phi$ for $x(t) = A \sin(\omega_0 t + \phi)$ is $0$ because at $t=0$,$x=0$ and $v > 0$. Therefore,the statement that the initial phase is $\pi$ is incorrect.

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