50.0 g stone is attached to the bottom of a vertical spring and set vibrating. The maximum speed of the stone is 15.0 cm / s and the period is 0.500 s.

(a) Find the spring constant of the spring.

(b) Find the amplitude of the motion.

(c) Find the frequency of oscillation.

i) Mass of the stone,$m=50.0\mathrm{gm}\mathrm{or}0.50\mathrm{kg}$ .

ii) Maximum speed of the oscillations,$v_{max}=15.0\mathrm{cm}/\mathrm{s}\mathrm{or}0.15\mathrm{m}/\mathrm{s}$ .

iii) Period of oscillations,$T=0.500s$ .

The vibrations of the stone attached to the spring follow simple harmonic oscillations. Hence, using the given formulae, we can get the spring constant, frequency, and amplitude of the oscillations.

Formula:

The angular frequency of the oscillations,$\omega =\frac{2\tau \tau }{T}$ (i)

The wavenumber of the oscillations,$k=m{\omega }^2$ (ii)

The maximum velocity of the wave,$v_{max}=\omega x_m$ (iii)

The frequency of the wave, $f=\frac{1}{T}$ (iv)

Using equation (i) and the given values, the angular frequency of the oscillations is given by:

$$\begin{gathered} \omega =\frac{2\times 3.14}{0.500} \\ =12.56\mathrm{rad}/\mathrm{s} \end{gathered}$$

Now, the spring constant using equation (ii) and the given values can be determined as:

$$\begin{gathered} k=0.050\times {\left(12.56\right)}^2 \\ =7.9\mathrm{N}/\mathrm{m} \end{gathered}$$

Hence, the value of the spring constant is 7.9 N/m .

For SHM, the amplitude of the motion using equation (iii) and the given values is given as:

$$\begin{gathered} x_m=\frac{v_{max}}{\omega } \\ =\frac{0.15}{12.56} \\ =0.0119\mathrm{m} \\ =1.19\mathrm{cm} \end{gathered}$$

Hence, the value of the amplitude is 1.19 cm .