Q.Show that even as well as odd harmonics are present as overtones in the case of an air column vibrating in a pipe open at both the ends. A wheel of moment of inertia 1 kg.m$^2$ is rotating at a speed of 30 rad/s. Due to friction on the axis, it comes to rest in 10 minutes. Calculate the average torque of the friction. OR Explain the formation of stationary waves by analytical method. Show that nodes and antinodes are equally spaced in stationary waves. The radius of gyration of a body about an axis, at a distance of 0.4 m from its centre of mass is 0.5 m. Find its radius of gyration about a parallel axis passing through its centre of mass.

In an open-open pipe, all harmonics (even and odd) are present as overtones; separately, angular deceleration from friction is found via $\tau=I\alpha$.

Option — Harmonics in a pipe open at both ends

For an air column vibrating in a pipe open at both ends, both ends must be antinodes (since air is free to move there), with at least one node in between. The possible modes (allowed standing-wave patterns) have length $L$ equal to an integral number of half-wavelengths:

$L = n\frac{\lambda_n}{2}, \qquad n=1,2,3,\dots \quad\Rightarrow\quad \lambda_n = \frac{2L}{n}$

The corresponding frequency is

$f_n = \frac{v}{\lambda_n} = \frac{nv}{2L} = n f_1$

where $f_1 = v/2L$ is the fundamental frequency. As $n$ takes all integer values $1,2,3,4,\dots$, the frequencies produced are $f_1, 2f_1, 3f_1, 4f_1,\dots$ — i.e. all integral multiples of the fundamental. Since overtones are simply all the frequencies present above the fundamental, this pipe produces overtones at $2f_1$ (2nd harmonic, 1st overtone), $3f_1$ (3rd harmonic, 2nd overtone), $4f_1$ (4th harmonic, 3rd overtone), and so on — that is, both even harmonics ($2f_1,4f_1,\dots$) and odd harmonics ($3f_1,5f_1,\dots$) occur as overtones. (This is in contrast to a pipe closed at one end, which produces only odd harmonics.)

Option — Numerical

Wheel: $I = 1\ \text{kg·m}^2$, initial $\omega_0 = 30\ \text{rad/s}$, comes to rest ($\omega=0$) in $t=10\ \text{min}=600\ \text{s}$ due to friction.

Using $\omega = \omega_0 - \alpha t$ (deceleration):

$0 = 30-\alpha(600) \quad\Rightarrow\quad \alpha = \frac{30}{600} = 0.05\ \text{rad/s}^2$

Average frictional torque:

$\tau = I\alpha = 1\times0.05 = 0.05\ \text{N·m}$

— OR (alternative) —

Formation of stationary waves (analytical method): as derived from the superposition of two identical waves travelling in opposite directions,

$y_1 = A\sin(kx-\omega t), \qquad y_2 = A\sin(kx+\omega t)$

$y = y_1+y_2 = 2A\sin(kx)\cos(\omega t)$

Nodes occur where the amplitude factor $\sin(kx) = 0$, i.e. $kx = n\pi \Rightarrow x = n\lambda/2$ (spacing $\lambda/2$).

Antinodes occur where $\sin(kx) = \pm1$, i.e. $x = (2n+1)\lambda/4$ (also spaced $\lambda/2$ apart, midway between successive nodes).

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