91影视

• 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

Hooke鈥檚 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鈥檚 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鈥檚 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,

$T=2蟺\sqrt{\frac{M}{k}}$ (ii)

Here, m is mass of the pendulum.

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

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

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

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.