Problem — Hanging mass on a spring suspension

Problem

An $80\;\text{g}$ mass hangs from a spring attached to the ceiling. At equilibrium the spring stretches by $\Delta x_0 = 4{,}0\;\text{cm}$ relative to its natural length. Displacing the mass $3\;\text{cm}$ further down and releasing it, calculate the period and the maximum speed.

Solution

At equilibrium the elastic force balances the weight, $k\,\Delta x_0 = mg$:

$k = \frac{mg}{\Delta x_0} = \frac{0{,}08 \cdot 9{,}81}{0{,}04} \approx 19{,}6$

$\ev{k \approx 19{,}6\;\text{N/m}}$

Angular frequency and period:

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{19{,}62}{0{,}08}} = \sqrt{245{,}3} \approx 15{,}66\;\text{rad/s}, \quad T = \frac{2\pi}{\omega}$

$\ev{T \approx 0{,}40\;\text{s}}$

With amplitude $A = 0{,}03\;\text{m}$, the maximum speed is $v_{\max} = A\omega = 0{,}03 \cdot 15{,}66$:

$\ev{v_{\max} \approx 0{,}47\;\text{m/s}}$

Note: the weight only shifts the centre of oscillation; it changes neither $k$ nor $T$.

Links

Topics: Oscillations and simple harmonic motion Concepts: Simple harmonic motion · Elastic force and Hooke’s law · Period and frequency Objects: Spring