A massless spring with spring constant 19 N/m hangs vertically. A body of mass 0.20 kgis attached to its free end and then released. Assume that the spring was unstretched before the body was released.

1. How far below the initial position the body descends?
2. Find frequency of the resulting SHM.
3. Find amplitude of the resulting SHM.

• Spring constant is, k = 19 N/m.
• Mass attached to the spring is, m = 0.20 kg.
• The spring was unstretched before the body was released.

Using the spring force, we can calculate how far below the initial portion of the body descends and the amplitude of SHM. Using the formula for the frequency of SHM, we can find the frequency of oscillations of the spring.

Formula:

The spring force of the oscillations, F = kx (i)

The frequency of the oscillations,$f=\frac{1}{2\tau \tau }\sqrt{\frac{k}{m}}$ (ii)

Using equation (i), the position at which the body falls is given as:

$$\begin{gathered} x=\frac{mg}{k}(\because F=mg) \\ =\frac{0.20\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2}{19\mathrm{N}/\mathrm{m}} \\ =0.103\mathrm{m} \end{gathered}$$

But when the body is released, it falls through distance x but continues equally far on the other side through the equilibrium position.

Total descent of the body is,

$$\begin{gathered} 2x=2\times 0.1032\mathrm{m} \\ =0.21\mathrm{m} \end{gathered}$$

Hence, the position at which the body descends is 0.21 m.

Using equation (ii), the frequency of the oscillations is given as:

$$\begin{gathered} f=\frac{1}{2\times 3.14}\sqrt{\frac{19}{0.20}} \\ =\frac{1}{6.2832}\times 9.7468 \\ =1.6\mathrm{Hz} \end{gathered}$$

Hence, the value of the frequency is 1.6 Hz.

Amplitude is the maximum displacement from the equilibrium and it is given as:

$$\begin{gathered} x_m=x \\ =0.10\mathrm{m} \end{gathered}$$

Hence, the value of the amplitude is 0.10 m.