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Some times in a movie we see that when an opera singer sings, the glass in her hand shatters. Also at times when we close the door, the window panes in the room start rattling. The phenomenon governing the above events is resonance. When the frequency of the opera singer (or closing door) matches with the frequency of the glass (or window), resonance occurs. In this demonstration we show this phenomenon of resonance.


  1. Tie one end of a thread to the nut and other end to a pencil to make a simple pendulum.
  2. Apply a small force on the pendulum to make it oscillate. Leave it free so that it starts oscillating with its natural frequency.
  3. Now rotate the pencil in clockwise and anticlockwise direction periodically with your fingers. Do it very slowly (at frequency much less than natural frequency of pendulum). See that the amplitude of the oscillations becomes very less.
  4. Again give a rotation to the pencil in a similar manner, but this time do it very fast (at frequency much more than the natural frequency of pendulum). The amplitude of oscillation is still very less.
  5. Now rotate the pencil in such a manner that frequency of the periodic rotation given to the pencil matches with the frequency of the pendulum. This will require some practice. See that the amplitude of oscillation of the pendulum becomes very large.

Every body oscillates with its natural frequency when it is left free. This natural frequency depends on the shape, size and material of the body. If the body is subjected to a periodic force the body starts oscillating with the frequency of this periodic force after some time. But the amplitude of these oscillations is very small if the forced frequency is different from the natural frequency. If the forced frequency happens to match with the natural frequency, the body gains a lot of energy and its amplitude become very large. We call this phenomenon as resonance.

Problems from IIT JEE

Problem (IIT JEE 1985): An air column in a pipe, which is closed at one end, will be in resonance with a vibrating tuning fork of frequency 264 Hz, if the length of the column is, (Speed of sound = 330 m/s.)

  1. 31.25 cm
  2. 62.50 cm
  3. 93.75 cm
  4. 125 cm

Solution: Natural frequencies of a pipe of length $l$, closed at one end, are given by, $\nu=nv/(4l)$, where $v={330}\;\mathrm{m/s}$ is the speed of sound and $n$ is an odd integer. Thus, the possible length of pipe (closed at one end) that can resonate with a tuning fork of frequency $\nu={264}\;\mathrm{Hz}$ are, \begin{align} l_n=\frac{nv}{4\nu}=\frac{n(330)}{4(264)}=0.3125 n\;m, \end{align} which gives $l_1={31.25}\;\mathrm{cm}$, $l_3={93.75}\;\mathrm{cm}$, $l_5={156.25}\;\mathrm{cm}$ etc. Hence, correct options are A and C.


JEE Physics Solved Problems in Mechanics