Question

A spring executes S.H.M. with mass 10 kg attached to it. The force constant of the spring is 10 N/m. If at any instant its velocity is 40 cm/s, the displacement at that instant is (Amplitude of S.H.M. = 0.5 m)

• A. 0.3 m
• B. 0.2 m
• C. 0.4 m
• D. 0.45 m

Hint:

In S.H.M., the velocity and displacement are related through the amplitude and angular velocity. Use the equation \( v = \omega \sqrt{A^2 - x^2} \) to find the displacement when velocity is known.

Answer 1

The Correct Option is A

Step 1: Using the equation of motion for S.H.M.
The equation of motion for simple harmonic motion is given by: \[ v = \omega \sqrt{A^2 - x^2} \] where \( v \) is the velocity, \( \omega \) is the angular velocity, \( A \) is the amplitude, and \( x \) is the displacement. We are given \( A = 0.5 \, \text{m} \), \( v = 40 \, \text{cm/s} = 0.4 \, \text{m/s} \), and the spring constant \( k = 10 \, \text{N/m} \). The angular velocity \( \omega \) is related to the spring constant and mass by: \[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{10}{10}} = 1 \, \text{rad/s} \] Step 2: Solving for displacement.
Now we substitute the values into the equation: \[ 0.4 = 1 \sqrt{(0.5)^2 - x^2} \] Squaring both sides: \[ 0.16 = 0.25 - x^2 \] Solving for \( x^2 \): \[ x^2 = 0.25 - 0.16 = 0.09 \] Thus: \[ x = \sqrt{0.09} = 0.3 \, \text{m} \] Step 3: Conclusion.
Thus, the displacement at the instant is 0.3 m, which is option (A).