CCommons

Damped and forced oscillations: does amplitude really go infinite at resonance, and which frequency does the system oscillate at?

I made a colour-coded summary but the resonance part doesn't add up.

Ananya Reddy11th Sample

15 Sept 2026 · 1 min read

I've finished the SHM part and now I'm on damped and forced oscillations. My notes have a green box for damped and an orange box for forced, but I'm unsure about several lines in them.

For damping, NCERT gives (for small damping)

x(t)=Ae−bt/2mcos⁡(ω′t+ϕ),ω′=km−b24m2x(t) = Ae^{-bt/2m}\cos(\omega' t + \phi), \qquad \omega' = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}}

So I understand the amplitude decays. But my energy line says E=12kA2e−bt/mE = \frac12 kA^2 e^{-bt/m}, and I can't tell if that is right or if I wrote the exponent wrongly. Also is ω′\omega' bigger or smaller than ω0=k/m\omega_0 = \sqrt{k/m}?

For forced oscillations the book says amplitude is maximum when the driving frequency ωd\omega_d equals the natural frequency ω\omega, and calls it resonance. So at resonance the amplitude is... infinite? My friend said yes, which sounds dangerous. Also, once I push a swing, does it swing with its own natural frequency or with my pushing frequency?

I'd love a clear short answer for each. Thanks!

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  • [+]Kabir Joshi11thSample· 1mo ago

    Starting from the amplitude formula in NCERT for the driven oscillator:

    A=F0m2(ω2−ωd2)2+ωd2b2A = \frac{F_0}{\sqrt{m^2(\omega^2-\omega_d^2)^2 + \omega_d^2 b^2}}

    At ωd=ω\omega_d = \omega the first term inside the root vanishes and A=F0ωdbA = \frac{F_0}{\omega_d b}. It's infinite only if b=0b = 0, i.e. no damping at all. In any real system b>0b > 0 so the amplitude is large but finite, and smaller damping gives a taller, sharper peak. Your energy line is right too: energy goes as amplitude squared, so A2∝e−bt/mA^2 \propto e^{-bt/m}.

  • [+]Rohan Sharma11thSample· 1mo ago

    resonance = amplitude infinite, bridge collapses (the famous soldiers-marching one, we saw the video in class). and with damping the frequency becomes more than the natural one because damping "pushes" it, so ω′>ω0\omega' > \omega_0. I think 😅 pls correct me if I'm wrong, I mostly remember the video part

    • [+]Farhan ShaikhDropperSample· 1mo ago

      Rohan, half of this is wrong. Look at the formula Ananya typed: you subtract b2/4m2b^2/4m^2 under the root, so ω′<ω0\omega' < \omega_0. Damping slows the oscillation a bit, the period gets slightly longer. For light damping the change is tiny so people say 'about the same'.

      And amplitude is not infinite, it is just maximum. Bridges and glasses break because the amplitude gets large enough to exceed what the material can take, not because it's infinite.

      • [+]Ananya Reddy11thSample· 1mo agoAuthor

        ahh okay so my energy line was right but my "ω′ greater" guess would have been wrong too. fixing the orange box now. thank you everyone 💛

  • [+]Kavya NaikDropperSample· 1mo ago

    Swing: think of the swing as having its own favourite rhythm. When you start pushing, it is confused for a while (transient) but after a short time it settles and swings at the rhythm you are pushing, not its own. That is the steady state of forced oscillation: frequency = driving frequency. Only when your pushing rhythm matches the swing's natural rhythm do all your pushes add up in phase, and the amplitude builds up. Amma used to push me at random on the swing and I went nowhere, the one time she timed it properly I almost flew off. 😄

  • [+]Ishita GoyalDropperSample· 1mo ago

    Quick summary I keep on my planner:

    • Free damped: amplitude ∝e−bt/2m\propto e^{-bt/2m}, energy ∝e−bt/m\propto e^{-bt/m}, ω′<ω0\omega' < \omega_0.
    • Forced steady state: oscillates at ωd\omega_d, not ω0\omega_0.
    • Resonance when ωd≈ω0\omega_d \approx \omega_0: amplitude is max, but finite if b>0b>0.
    • More damping: peak lower and broader.

    NEET likes 'a system oscillating under periodic force: its frequency is that of the ___' - the answer is the driving force.

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