Question

Under the action of a force \( F = -75y \) where F is in Newton and y is in meters, an object of mass 3 kg executes simple harmonic motion. If the velocity of the object at the mean position is 2.5 ms\(^{-1} \), the maximum acceleration of the object is

• A. \( 5 \) ms\(^{-2} \)
• B. \( 7.5 \) ms\(^{-2} \)
• C. \( 10 \) ms\(^{-2} \)
• D. \( 12.5 \) ms\(^{-2} \)

Hint:

Relate the given force to the acceleration using Newton's second law. Compare the acceleration expression with the standard form for SHM to find the angular frequency. Use the maximum velocity at the mean position to determine the amplitude of the SHM. Finally, calculate the maximum acceleration using the angular frequency and amplitude.

Answer 1

The Correct Option is D

The force acting on the object is \( F = -75y \).
According to Newton's second law, \( F = ma \), where \( m \) is the mass and \( a \) is the acceleration.
So, \( ma = -75y \).
The acceleration of the object is \( a = -\frac{75}{m} y \).
Given mass \( m = 3 \) kg, the acceleration is \( a = -\frac{75}{3} y = -25y \).
For simple harmonic motion, the acceleration is given by \( a = -\omega^2 y \), where \( \omega \) is the angular frequency.
Comparing the two expressions for acceleration, we have \( \omega^2 = 25 \), so \( \omega = 5 \) rad/s.
The velocity of the object in SHM is given by \( v = \omega \sqrt{A^2 - y^2} \), where \( A \) is the amplitude of the motion.
At the mean position, \( y = 0 \), and the velocity is maximum, \( v_{max} = \omega A \).
Given that the velocity at the mean position is \( v_{max} = 2.
5 \) ms\(^{-1} \).
So, \( 2.
5 = 5 A \).
The amplitude \( A = \frac{2.
5}{5} = 0.
5 \) m.
The maximum acceleration in SHM occurs at the extreme positions (\( y = \pm A \)) and is given by \( a_{max} = \omega^2 A \).
Substituting the values of \( \omega^2 \) and \( A \): \( a_{max} = (25)(0.
5) = 12.
5 \) ms\(^{-2} \).