Question

When a 20 Ncan is hung from the bottom of a vertical spring, it causes the spring to stretch 20 cm .

1. What is the spring constant?
2. This spring is now placed horizontally on a frictionless table. One end of it is held fixed, and the other end is attached to a 5.0 Ncan. The can is then moved (stretching the spring) and released from rest. What is the period of the resulting oscillation?

Short answer

1. The value of the spring constant is $1.0\times {10}^2\mathrm{N}/\mathrm{m}$
2. The period of oscillation of the system is 0.45 s

Step-by-step solution

The given data

• The weight of the can is,$w=20N$
• The displacement of the spring is, x=20 cm or 0.20 m
• Weight of the another can is, W=5N

Understanding the concept of SHM

Hooke’s law states that when the pendulum is displaced from its equilibrium position, it is acted upon by a restoring force which is directly proportional to the displacement of the pendulum and always acts towards the equilibrium position.

Using Hooke’s law, we can find the value of the spring constant. Then using the formula for the period of oscillation for S.H.M we can find the period of oscillation of spring.

Formulae:

The restoring force of a spring, by Hooke’s law,

$F=-Kx$ (i)

Here, F is restoring force, k is force constant, and x is displacement from the mean position.

The period of oscillation in S.H.M,

$T=2\pi \sqrt{\frac{M}{k}}$ (ii)

Here, m is mass of the pendulum.

a) Calculation of the spring constant

Applying equation (i) to the given system, we can get the spring constant as:

$$\begin{gathered} w-kx=0 \\ k=\frac{w}{x} \\ k=\frac{20N}{0.20m} \\ k=1.0\times {10}^{2}N/m \end{gathered}$$

Therefore, the value of the spring constant is $k=1.0\times {10}^{2}N/m$

b) Calculation of the period of oscillations

The mass of the 5N can is given as:

$\begin{array}{l}M=\frac{W}{g}\\ =\frac{5N}{9.8m/s^2}\\ =0.51kg\end{array}$

The period of oscillation of system is calculated by substituting the given values in equation (ii) as,

role="math" localid="1660979528272" $$\begin{gathered} T=2(3.142)\sqrt{\frac{0.51kg}{100N/m}} \\ =0.45s \end{gathered}$$

Therefore, the period of oscillation of system is 0.45 s.