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 ______
Correct Answer: 6
Solution and Explanation
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.
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Concepts Used:
Oscillations
Oscillation is a process of repeating variations of any quantity or measure from its equilibrium value in time . Another definition of oscillation is a periodic variation of a matter between two values or about its central value.
The term vibration is used to describe the mechanical oscillations of an object. However, oscillations also occur in dynamic systems or more accurately in every field of science. Even our heartbeats also creates oscillations. Meanwhile, objects that move to and fro from its equilibrium position are known as oscillators.
Read More: Simple Harmonic Motion
Oscillation- Examples
The tides in the sea and the movement of a simple pendulum of the clock are some of the most common examples of oscillations. Some of examples of oscillations are vibrations caused by the guitar strings or the other instruments having strings are also and etc. The movements caused by oscillations are known as oscillating movements. For example, oscillating movements in a sine wave or a spring when it moves up and down.
The maximum distance covered while taking oscillations is known as the amplitude. The time taken to complete one cycle is known as the time period of the oscillation. The number of oscillating cycles completed in one second is referred to as the frequency which is the reciprocal of the time period.