Question

A 2.00 kgblock hangs from a spring. A 300 kgbody hung below the block stretches the spring 2.00 cmfarther.

1. What is the spring constant?
2. If the 300 kgbody is removed and the block is set into oscillation, find the period of the motion.

Short answer

a) The spring constant is 147 N/m .

b) The period of motion is 0.733 s .

Step-by-step solution

The given data

• Mass of the block,m=2.00 kg .
• Mass of the body,$m_b=300\mathrm{gm}\mathrm{or}0.30\mathrm{kg}$.
• The extension of the spring,x=2.00 cm or 0.02 m .

Understanding the concept of SHM

Hooke’s law states that the restoring force is directly proportional to the displacement of the oscillating body and acts in the opposite direction to the displacement. The extra weight (body) attached to the string stretches the spring. The spring obeys Hooke’s law and its motion exhibits simple harmonic motion.

Formula:

The stretched force applied on a body,$F=-kx$ (i)

The period of oscillations,$T=\frac{2\tau \tau }{\omega }$ (ii)

The angular frequency of an oscillation, $\omega =\sqrt{\frac{k}{m}}$ (iii)

a) Calculation of spring constant

The body attached to the block stretches the spring by amount x. The spring obeys Hooke’s law. Hence, we can write,

$$\begin{gathered} F=\mathrm{Weight}\mathrm{of}\mathrm{the}\mathrm{body} \\ ={\mathrm{m}}_{\mathrm{b}}\mathrm{g} \\ =0.30\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

So, considering the magnitude only using equation (i), we get the spring constant as:$$\begin{gathered} k=\frac{F}{x} \\ =\frac{0.30\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2}{0.02\mathrm{m}} \\ =147\mathrm{N}/\mathrm{m} \end{gathered}$$

Hence, the value of spring constant is 147 N/m .

b) Calculation of period of oscillations

Using equations (ii) and (iii), we get the period of oscillations as:

$$\begin{gathered} T=2\tau \tau \sqrt{\frac{m}{k}} \\ =2\times 3.14\times \sqrt{\frac{2.00kg}{147N/m}} \\ =0.733s \end{gathered}$$

Hence, the value of period is 0.733 s .