Question

A spring with spring constant $15\mathrm{N}/\mathrm{m}$hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down $6.0\mathrm{cm}$and released. If the ball makes $30$oscillations in localid="1650030754517" $20\mathrm{s}$, what are its localid="1650030759248" $\left(a\right)$mass and localid="1650030763230" $\left(b\right)$maximum speed?

Short answer

The mass is$m=170g$and maximum speed is$v_{max}=56.5cm/s$.

Step-by-step solution

Given data.

Given

• $k=15N/m$
  
• $A=6.0cm$
  
• $N=30$ oscillations
• $t_N=20S$
  

Required $a)$$m$and $b)v_{max}$

To find mass.

$a)$The equation connects between $k,\omega ,m$is given by

${\omega }^2=\frac{k}{m}\Rightarrow m=\frac{k}{{\omega }^2}\Rightarrow \left(1\right)$

The angular frequency is given by

$\omega =\frac{2\pi }{T}\Rightarrow \left(2\right)$

Where $T$is the periodic time.

Periodic time is the time required to complete 1 oscillation so periodic time is

$T=\frac{t_N}{N}=\frac{2}{3}$

Substitution in $\left(2\right)$to get that

$\omega =\frac{2\pi }{2/3}=3\pi$

So substitution in $\left(2\right)$yields

$m=\frac{15}{(3\pi {)}^2}\approx 0.170\mathrm{kg}=170\mathrm{g}$

To find maximum speed

b) The equation of velocity is

$v\left(t\right)=-v_{\max }\sin \left(\omega t+{\varphi }_o\right)\Rightarrow \left(3\right)$

But velocity is the derivative of$x\left(t\right)$which is given by

$v\left(t\right)=-A\omega \sin \left(\omega t+{\varphi }_o\right)\Rightarrow \left(4\right)$

Compare $\left(3\right)$and $\left(4\right)$to get that

$v_{\max }=A\omega =3\pi \times 6\approx 56.5\mathrm{cm}/\mathrm{s}$