A tray of mass $12 \,kg$ is supported by two identical springs as shown in figure. When the tray is pressed down slightly and then released, it executes SHM with a time period of $1.5\,s$. The spring constant of each spring is
- $50 \,Nm^{-1}$
- $0$
- $105 \,Nm^{- 1}$
- $\infty$
The Correct Option is C
Approach Solution - 1
The time period \(T\) for simple harmonic motion (SHM) for a system with springs in parallel can be given by the formula: \[ T = 2\pi \sqrt{\frac{m}{k_{\text{eff}}}} \] where \(m\) is the mass, and \(k_{\text{eff}}\) is the effective spring constant for the system. For two identical springs in parallel, the effective spring constant is: \[ k_{\text{eff}} = 2k \] Given that the time period \(T = 1.5\) s and \(m = 12\) kg, we can solve for \(k\) as follows: \[ 1.5 = 2\pi \sqrt{\frac{12}{2k}} \] Squaring both sides: \[ 2.25 = 4\pi^2 \frac{12}{2k} \] Simplifying: \[ 2.25 = 24\pi^2 \frac{1}{k} \] Solving for \(k\): \[ k = \frac{24\pi^2}{2.25} \approx 105 \, \text{N/m} \] Thus, the spring constant of each spring is approximately 105 N/m.
Therefore, the correct answer is (C) 105 N/m.
Approach Solution -2
\(T=2π \sqrt{\frac{m}{k_{eff}}} \)
\(\frac{3}{2}=2π \sqrt{\frac{12}{2k}} \)
\(\Rightarrow = 4 \pi^{2} \times \frac{12}{2k} \approx 105Nm^{-1}\)
Approach Solution -3
For a system of springs in parallel, the effective spring constant \( k_{\text{eff}} \) is the sum of the spring constants of the individual springs. Since the springs are identical, the effective spring constant is: \[ k_{\text{eff}} = 2k \] where \( k \) is the spring constant of each individual spring. The time period \( T \) for simple harmonic motion (SHM) of a mass \( m \) attached to a spring with spring constant \( k_{\text{eff}} \) is given by: \[ T = 2\pi \sqrt{\frac{m}{k_{\text{eff}}}} \] Substitute the known values: \[ 1.5 = 2\pi \sqrt{\frac{12}{2k}} \] Solve for \( k \): \[ 1.5 = 2\pi \sqrt{\frac{12}{2k}} \quad \Rightarrow \quad \frac{1.5}{2\pi} = \sqrt{\frac{12}{2k}} \quad \Rightarrow \quad \left( \frac{1.5}{2\pi} \right)^2 = \frac{12}{2k} \] \[ \frac{1.5^2}{(2\pi)^2} = \frac{12}{2k} \quad \Rightarrow \quad \frac{2.25}{4\pi^2} = \frac{12}{2k} \quad \Rightarrow \quad \frac{2.25}{4\pi^2} \times 2k = 12 \] \[ k =105 \, \text{N/m} \]
Thus, the spring constant of each spring is \( 105 \, \text{N/m} \).
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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.