Question

A block of 50 g mass is connected to a spring of spring constant 500 Nm\(^{-1}\). It is extended to the maximum and released. If the maximum speed of the block is 3 ms\(^{-1}\), then the length of extension is:

• A. 4 cm
• B. 1 cm
• C. 2.5 cm
• D. 3 cm
• E. 5 cm

Hint:

For simple harmonic motion, the maximum speed is given by \( v_{{max}} = A \omega \). The amplitude can be found by dividing the maximum speed by the angular frequency \( \omega = \sqrt{\frac{k}{m}} \).

Answer 1

The Correct Option is D

The motion of the block is simple harmonic motion (SHM), and the maximum speed of the block is given by: \[ v_{{max}} = A \omega \]
where \( A \) is the amplitude (maximum extension), and \( \omega \) is the angular frequency of the SHM.
The angular frequency \( \omega \) is related to the spring constant \( k \) and the mass \( m \) of the block by the equation: \[ \omega = \sqrt{\frac{k}{m}}. \] Given: - \( k = 500 \, {N/m} \), - \( m = 50 \, {g} = 0.05 \, {kg} \), - \( v_{{max}} = 3 \, {ms}^{-1} \). First, calculate \( \omega \): \[ \omega = \sqrt{\frac{500}{0.05}} = \sqrt{10000} = 100 \, {rad/s}. \] Now, using the maximum speed equation: \[ v_{{max}} = A \omega, \] substitute the known values: \[ 3 = A \times 100 \quad \Rightarrow \quad A = \frac{3}{100} = 0.03 \, {m}. \]
Thus, the length of extension \( A \) is \( 0.03 \, {m} = 3 \, {cm} \).
Thus, the correct answer is option (D), 3 cm.