Question

The figure shows a block of mass \(m = 20\) kg attached to a pair of identical linear springs, each having a spring constant \(k = 1000\) N/m. The block oscillates on a frictionless horizontal surface. Assuming free vibrations, the time taken by the block to complete ten oscillations is ............... seconds. (Rounded off to two decimal places) Take \(\pi = 3.14\). 

Hint:

Recognizing spring configurations is crucial. Springs are in {parallel} if they experience the same displacement. They are in {series} if they experience the same force. In this common setup with a mass between two fixed walls, the springs are in parallel because moving the mass by \(\Delta x\) stretches one and compresses the other by \(\Delta x\).

Answer 1

Step 1: Understanding the Concept:
The problem describes a simple harmonic oscillator. The block is connected to two springs. Since a displacement of the mass to the right will cause one spring to compress and the other to extend, both springs will exert a restoring force in the same direction (to the left). This configuration means the springs are acting in parallel. We need to find the equivalent spring constant, then the natural frequency and period of oscillation.
Step 2: Key Formula or Approach:
1. Equivalent Spring Constant (\(k_{eq}\)): For springs in parallel, the equivalent stiffness is the sum of individual stiffnesses: \(k_{eq} = k_1 + k_2\).
2. Natural Frequency (\(\omega_n\)): The angular frequency of a spring-mass system is given by \(\omega_n = \sqrt{\frac{k_{eq}}{m}}\).
3. Time Period (\(T\)): The time period for one full oscillation is \(T = \frac{2\pi}{\omega_n}\).
4. Total Time: The time for \(N\) oscillations is \(t = N \times T\).
Step 3: Detailed Calculation:
Given:
- Mass, \(m = 20\) kg
- Spring constant, \(k_1 = k_2 = k = 1000\) N/m
- Number of oscillations, \(N = 10\)
- \(\pi = 3.14\)
1. Calculate the equivalent spring constant: \[ k_{eq} = k + k = 1000 + 1000 = 2000 \text{ N/m} \] 2. Calculate the natural angular frequency: \[ \omega_n = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{2000 \text{ N/m}}{20 \text{ kg}}} = \sqrt{100} = 10 \text{ rad/s} \] 3. Calculate the time period for one oscillation: \[ T = \frac{2\pi}{\omega_n} = \frac{2\pi}{10} = \frac{\pi}{5} \text{ s} \] 4. Calculate the total time for ten oscillations: \[ t = 10 \times T = 10 \times \frac{\pi}{5} = 2\pi \text{ s} \] 5. Substitute the value of \(\pi\): \[ t = 2 \times 3.14 = 6.28 \text{ s} \] Step 4: Final Answer:
The time taken by the block to complete ten oscillations is 6.28 seconds.
Step 5: Why This is Correct:
The steps correctly identify the springs as being in parallel, calculate the equivalent stiffness, and then use the standard formulas for natural frequency and time period of a simple harmonic oscillator to find the required time for ten oscillations.