Question

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.

Short answer

(a) The spring constant of the spring is 7.9 N/m .

(b) The amplitude of the motion is 1.19 cm .

(c) The frequency of the motion is 2.00Hz .

Step-by-step solution

The given data

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,$T=0.500s$ .

Understanding the concept of SHM

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)

(a) Calculation of the spring constant

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 .

b) Calculation of amplitude of the motion

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 .