Adding a constant term to the characteristic equation of SHM, will it be SHM or not

 Great question — this gets right to the definition of Simple Harmonic Motion (SHM).

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1. Standard SHM equation

In SHM, the equation is:

$$\frac{d^2 x}{dt^2} = -\omega^2 x$$

This means:

• Acceleration is directly proportional to displacement
• And always towards the equilibrium position (x = 0)

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2. Modified equation with a constant term

Now suppose we have:

$$\frac{d^2 x}{dt^2} = -\omega^2 x + C$$

This extra constant term (C) changes things.

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3. Is it still SHM?

👉 Strict answer: No (not in the standard form)

Because:

• Acceleration is not purely proportional to displacement from x = 0
• There is an additional constant force

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4. Key insight: Shift of equilibrium

We can rewrite the equation:

$$\frac{d^2 x}{dt^2} = -\omega^2 \left(x - \frac{C}{\omega^2}\right)$$ \[a=-\omega^2\left(x-x_o\right) \]

Equilibrium shifted to :

\[x_o=\frac{C}{\omega^2}\] $$x' = x - \frac{C}{\omega^2}$$

Then:

$$\frac{d^2 x'}{dt^2} = -\omega^2 x'$$

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5. Final conclusion

✅ Yes — it is still SHM, but about a shifted equilibrium position

Differences from usual SHM:

Feature	Usual SHM	Modified system
Equilibrium position	$x=0$	$x = \frac{C}{\omega^2}$

Restoring force          $-\omega^2 x$$-\omega^2 (x - x_{\text{o}})$
Nature of motion          SHM        SHM (shifted origin) 

 Angular frequency      $\omega$                Same $\omega$

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6. Physical interpretation

This situation appears when:

• A constant external force is applied
  (e.g., gravity + spring system)

Example:

• Vertical spring:
  • Gravity adds constant term
  • New equilibrium shifts downward
  • Oscillations still SHM about new point

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7. One-line takeaway

👉 Adding a constant term does NOT destroy SHM — it just shifts the equilibrium position.

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JEE / NEET Level Problems

Q1. A particle obeys the equation:

\[\frac{d^2x}{dt^2}=-4x+8\]

Find equilibrium position and time period.

Solution:

Compare with:

\[a=-\omega^2 (x -x_o)\]

\[a=-4(x-2)\]

\[\omega^2 =4 \rightarrow \omega = 2 rad/s \]

Equilibrium position:

\[x_o= 2\]

Time period:

\[T=\frac{2\pi}{\omega}=\pi \sec \]

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Q2. A mass-spring system is subjected to a constant force F. Show that motion is SHM and find new equilibrium.

Solution:

Equation:

\[m \frac{d^2x}{dt^2}=-kx+F\]

\[\frac{d^2x}{dt^2}=-\frac{k}{m}x+\frac{F}{m}\]

Compare:

\[a=-\omega^2 (x -x_o)\]

\[a=-\frac{k}{m} (x-\frac{F}{k})\]

\[\omega^2=\frac{k}{m}, \ x_o=\frac{F}{k} \]

Equilibrium: Thus motion is SHM about shifted position x_o.

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Q3. A particle oscillates under:

\[\frac{d^2x}{dt^2}=-9x+18\]

Find maximum velocity if amplitude about new equilibrium is 3 m.

Solution:

\[\omega = 3 \hspace{0.25cm} rad/s \]

\[v_{max}=A\omega= 3 \times 3 = 9 \hspace{0.25cm}m/s \]