Question

On a frictionless horizontal plane, a bob of mass $m=01\, kg$ is attached to a spring with natural length $l_0=01 \,m$. The spring constant is $k_1=0009 \,Nm ^{-1}$ when the length of the spring $l>l_0$ and is $k_2=0016\, Nm ^{-1}$ when $l < l_0$. Initially the bob is released from $l=015 \,m$. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T =(n \pi) s$, then the integer closest to $n$ is ______

Answer 1

Correct Answer: 6

1. Given Information:
The mass of the bob is \( m = 0.1 \, \text{kg} \),
The natural length of the spring is \( l_0 = 0.1 \, \text{m} \),
The spring constant when the spring is stretched beyond its natural length (\( l > l_0 \)) is \( k_1 = 0.09 \, \text{N/m} \),
The spring constant when the spring is compressed (\( l < l_0 \)) is \( k_2 = 0.016 \, \text{N/m} \),
The initial displacement of the bob is \( l = 0.15 \, \text{m} \),
We need to find the time period \( T \) of the oscillation.

2. Understanding Hooke's Law:
Hooke's law states that the force exerted by a spring is proportional to the displacement from its natural length. The general form of Hooke's law is:

\[ F = -k \Delta l \] where:
\( F \) is the force,
\( k \) is the spring constant,
\( \Delta l \) is the displacement from the natural length.

3. Determining the Force and the Restoring Force:
In this case, the spring is stretched by \( 0.15 \, \text{m} - 0.1 \, \text{m} = 0.05 \, \text{m} \), so \( \Delta l = 0.05 \, \text{m} \). The restoring force acting on the bob is provided by Hooke’s law for \( l > l_0 \), where \( k = k_1 = 0.09 \, \text{N/m} \). Therefore, the force is:

\(F = -k_1 \times 0.05 = -0.09 \times 0.05 = -0.0045 \, \text{N}\).

4. Equation of Motion and Time Period:
For simple harmonic motion (SHM), the time period \( T \) of the oscillation is given by the formula:

\[ T = 2 \pi \sqrt{\frac{m}{k}} \] where: - \( m \) is the mass of the bob, - \( k \) is the effective spring constant.

For \( l > l_0 \), we use \( k = k_1 = 0.09 \, \text{N/m} \) to calculate the time period of oscillation. Substituting the values:

\(T = 2 \pi \sqrt{\frac{0.1}{0.09}} = 2 \pi \sqrt{\frac{10}{9}} \approx 2 \pi \times 1.054 \approx 6.621 \, \text{seconds}.\)

5. Closest Integer:
Since the time period is approximately \( 6.621 \, \text{s} \), the integer closest to \( n \) is 6.

Final Answer:
The integer closest to \( n \) is 6.