Question

State the differential equation of linear S.H.M. Hence, obtain expression for:
(i) Acceleration
(ii) Velocity
Correct Answer: The differential equation of linear simple harmonic motion (S.H.M.) is \( m \frac{d^2x}{dt^2} = -kx \). From this, we can derive the expressions for acceleration and velocity.

Hint:

In simple harmonic motion, acceleration and velocity are related to displacement through the angular frequency \( \omega \), with acceleration being proportional to displacement and velocity being phase-shifted by \( \frac{\pi}{2} \).

Answer 1

Step 1: Differential Equation of S.H.M.
The general form of the equation for linear simple harmonic motion is: \[ m \frac{d^2x}{dt^2} = -kx \] where \( m \) is the mass, \( k \) is the spring constant, and \( x \) is the displacement of the particle.
Step 2: Acceleration.
The acceleration \( a \) is the second derivative of displacement with respect to time: \[ a = \frac{d^2x}{dt^2} \] From the differential equation, we know that: \[ m \frac{d^2x}{dt^2} = -kx \] Therefore, the acceleration is: \[ a =
- \frac{k}{m} x \] This shows that the acceleration is directly proportional to the displacement but in the opposite direction, characteristic of simple harmonic motion.
Step 3: Velocity.
The velocity \( v \) is the first derivative of displacement with respect to time: \[ v = \frac{dx}{dt} \] From the solution of the differential equation, we know the displacement as a function of time is: \[ x(t) = A \cos(\omega t + \phi) \] where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant. The velocity is: \[ v(t) = -A \omega \sin(\omega t + \phi) \] Thus, the velocity is proportional to the displacement, with a phase difference of \( \frac{\pi}{2} \).