Example Constant Gravity force added to oscillating spring force

Adding a constant term to the Characteristic Equation of SHM, will it remain as SHM Explanation

Instead of the Horizontal Spring oscillations, where only spring force is present, 

Suppose if we do vertical spring oscillations, where in addition to the spring force, constant gravitational force is acting always. How will the system behave? will it remain as SHM or not?

Simple Harmonic Motion with Constant Force (Gravity)

This section explains how a spring-mass system behaves when a constant force (like gravity) acts along with the restoring spring force. This is an important concept for CBSE, JEE, and NEET Physics.

1. Spring Force

\( F_s = -kx \)

This is the restoring force of the spring.

2. Equilibrium Condition

At equilibrium, net force is zero:

\( F_s = mg \)

\( -k x_0 = mg \)

\( k x_0 = mg \)

\( x_0 = \frac{mg}{k} \)

This shows that the equilibrium position shifts downward due to gravity.

3. Motion About Equilibrium

Let the displacement from natural length be \( x \).

\( F_s = -kx \)

Applying Newton’s Second Law:Restoring force

\( ma = -kx + mg \)

\( a = -\frac{k}{m} \left(x - \frac{mg}{k}\right) \)

Substitute \( x_0 = \frac{mg}{k} \):

\( a = -\frac{k}{m}(x - x_0) \)

4. Conclusion: SHM Condition

The motion is still Simple Harmonic Motion (SHM), but:

• Equilibrium position is shifted to \( x_0 = \frac{mg}{k} \)
• Acceleration depends on displacement from new equilibrium

\( a = -\omega^2 (x - x_0) \)

\( \omega = \sqrt{\frac{k}{m}} \)

5. Key Insight

Even when a constant force (like gravity) is present, the system still performs SHM.

The only change is a shift in equilibrium position. The angular frequency remains unchanged.

6. Important Points for Exams

• Gravity does NOT affect time period
• Time period: \( T = 2\pi \sqrt{\frac{m}{k}} \)
• SHM always occurs about equilibrium position