Question

The system shown in Fig. 14.17 is mounted in an elevator. What happens to the period of the motion (does it increase, decrease, or remain the same) if the elevator (a) accelerates upward at$\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ ; (b) moves upward at a steady $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$; (c) accelerates downward at $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$? Justify your answers.

Short answer

(a) Accelerates upward at$5.0\mathrm{m}/{\mathrm{s}}^2$ is remain the same.

(b) Moves upward at a steady$5.0\mathrm{m}/\mathrm{s}$ is remain the same.

(c) Accelerates downward at$5.0\mathrm{m}/{\mathrm{s}}^2$ is remain the same.

Step-by-step solution

Determine the formula for SHM

The period of a vertical mass-spring system is given by

$\mathbf{T}\mathbf{=}\mathbf{2}\mathbf{\tau \tau }\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}$

Determine the period of motion

The period of a vertical mass-spring system is given by

$\mathrm{T}=2\mathrm{\tau \tau }\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$

Where kis the spring force constant and m is mass.

Since the motion of the object is affected by the net acceleration of the object${\mathrm{a}}_{\mathrm{net}}=\mathrm{g}\pm \mathrm{a}$.

Here the period is independent of the acceleration due to gravity. Therefore the period of the motion if the elevator

(a) accelerates upward at$\mathrm{a}=5.0\mathrm{m}{\mathrm{s}}^{‐2}$ will remain the same.

(b) moves upward at a steady$\mathrm{a}=5.0\mathrm{m}{\mathrm{s}}^{‐1}$ will remain the same.

(c) accelerates downward at$\mathrm{a}=5.0\mathrm{m}{\mathrm{s}}^{‐2}$ will remain the same.