Additional constant term in the characteristic equation in SHM

 

SHM with Constant Term (Shifted Equilibrium)

In standard Simple Harmonic Motion (SHM), acceleration is proportional to displacement:

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

If a constant term is added:

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

This represents SHM about a shifted equilibrium position.

New Equilibrium Position

\[a=-\omega^2 (x -\frac{C}{\omega^2})\]

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

\[x_o =\frac{C}{\omega^2}\]

The system still performs SHM, but about this new position instead of zero.

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Graph Comparison

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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:

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

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

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Key Takeaways

• Constant term shifts equilibrium
• Motion remains SHM
• Angular frequency unchanged
• Oscillation occurs about new position