A spring with spring constant $15.0N/m$hangs from the ceiling.

A $500g$ball is attached to the spring and allowed to come to rest. It is then pulled down $6.0cm$and released. What is the time constant if the ball’s amplitude has decreased to $3.0cm$after $30$oscillations?

Damping oscillation: The mechanical energy E in a real oscillating system decreases during oscillation because external forces, such as resistance, prevent the oscillation and convert the mechanical energy into thermal energy. The real oscillator and its movement are then said to be damped. If the damping force gives $F=-bv$, where $v$is the speed of the oscillation and$b$ is the damping constant, then the displacement of the oscillation is given by

localid="1650087316439" $x\left(t\right)=A{e}^{-bt/2m}\cos (\omega t+\varphi )$

where localid="1650087334599" $\omega$, the angular frequency of the damped oscillator , is given by

localid="1650087323381" $\omega =\sqrt{\frac{g}{L}-\frac{b^2}{4m^2}}$

Here localid="1650087342562" $\surd (g/L)$can be the angular frequency of an undamped oscillationslocalid="1650087348719" $(b=0)$.

The period of oscillation in simple harmonic motion is given by

localid="1650087355497" $T=2\pi \sqrt{\frac{m}{k}}$

• The spring constant of the spring is: $k=15.0N/m$.
• The ball's mass can be:$m=\left(500g\right)\left(\right(1kg/1000g\left)\right)=0.500kg$ .
• The starting amplitude of the ball can be: $A_0=6.0\mathrm{cm}$.
• The amplitude of the ball after 30 oscillations is: $A_{30}=3.0\mathrm{cm}$.

The objective is to calculate the time constant.

According to equation $\left(1\right)$, the amplitude of the ball's oscillation is expressed as function of time follows

$A\left(t\right)=A_0{e}^{-bt/2m}=A_0{e}^{-t/2\tau }$

where$A_0$is the initial amplitude of oscillator and$\tau =m/b$is the time constant for loss of energy level. The amplitude after $30$oscillation is

$A_{30}=A_0{e}^{-30T/2\tau }$

$T$is the oscillation of the ball.

$\frac{A_{30}}{A_0}=e^{-30T/2\tau }$

$n\left(\frac{A_{30}}{A_0}\right)=-\frac{30T}{2\tau }$

$\tau =-\frac{30T}{2\ln \left(\frac{A_{30}}{A_0}\right)}$

Substitute $T$value

$\tau =-\frac{30\left(2\pi \sqrt{\frac{m}{k}}\right)}{2\ln \left(\frac{A_{30}}{A_0}\right)}$

$=-\frac{30\pi }{\ln \left(\frac{A_{30}}{A_0}\right)}\sqrt{\frac{m}{k}}$

Substituting numerical values

$\tau =-\frac{30\pi }{\ln \left(\frac{3.0\mathrm{cm}}{6.0\mathrm{cm}}\right)}\sqrt{\frac{0.500\mathrm{kg}}{15.0\mathrm{N}/\mathrm{m}}}$

$=25s$