Question

A spring with spring constant \( K \) stores 5 J energy when stretched by 25 cm. It was fixed at one end in vertical position and the other end is attached by mass \( m \). Its frequency is 5 oscillations/sec. Then the values of \( m \) and \( K \) are:

• A. \( m = 0.16 \, {kg}, K = 160 \, {N/m} \)
• B. \( m = 16 \, {kg}, K = 160 \, {N/m} \)
• C. \( m = 0.16 \, {kg}, K = 16 \, {N/m} \)
• D. \( m = 16 \, {kg}, K = 16 \, {N/m} \)

Hint:

Always ensure units are consistent and formulas are applied correctly for accuracy in physics problems involving energy and motion.

Answer 1

The Correct Option is A

We are given the following data: 
- The spring stores 5 J of energy when stretched by 25 cm (0.25 m). 
- The frequency of oscillation is 5 oscillations per second. 
Step 1: Calculate the spring constant \( K \) using the potential energy formula. 
The potential energy stored in a spring is given by: \[ E = \frac{1}{2} K x^2 \] where \( E \) is the energy stored in the spring, \( K \) is the spring constant, and \( x \) is the displacement from the equilibrium position. Given that the energy stored in the spring is 5 J and the displacement \( x = 0.25 \, {m} \), we can solve for \( K \): \[ 5 = \frac{1}{2} K (0.25)^2 \] \[ 5 = \frac{1}{2} K \times 0.0625 \] \[ K = \frac{5}{0.03125} = 160 \, {N/m} \] Step 2: Use the formula for the frequency of oscillation. The frequency \( f \) of oscillation for a mass-spring system is given by: \[ f = \frac{1}{2\pi} \sqrt{\frac{K}{m}} \] where \( f \) is the frequency, \( K \) is the spring constant, and \( m \) is the mass. We are given that the frequency is 5 oscillations/sec, so: \[ 5 = \frac{1}{2\pi} \sqrt{\frac{160}{m}} \] Squaring both sides: \[ 25 = \frac{1}{4\pi^2} \times \frac{160}{m} \] Solving for \( m \): \[ m = \frac{160}{4\pi^2 \times 25} = \frac{160}{4 \times 9.87 \times 25} \] \[ m = \frac{160}{987.5} \approx 0.16 \, {kg} \] Conclusion: The values of \( m \) and \( K \) are: \[ m = 0.16 \, {kg}, \, K = 160 \, {N/m} \] So, the correct answer is (1).