(a) What is a wave motion? The equation \(y = A \sin \frac{2\pi}{\lambda}(Vt-X)\) represents a wavetrain in which y is the vertical displacement of a partic...

(a)

Wave motion is the propagation of a disturbance from one point to another, transmitting energy without any net transfer of matter.

For the wave equation

\[y=A\sin\left[\frac{2\pi}{\lambda}(Vt-X)\right],\]

0 is the amplitude, that is, the maximum displacement of a particle of the medium from its equilibrium position.

\(\lambda\) is the wavelength, that is, the distance between two successive points vibrating in the same phase, such as two successive crests.

(b)(i)

Set up a resonance tube containing water and a movable vertical tube. Strike the sound source, such as a tuning fork, gently and hold it just above the open end of the tube. Raise the tube slowly so that the length \(l\) of the enclosed air column increases.

The sound becomes loudest at resonance. At the first resonance, the air column has a node at the water surface and an antinode at the open end; hence, neglecting end correction,

\[l=\frac{\lambda}{4}.\]

Therefore,

\[\lambda=4l\]

and, if \(v\) is the velocity of sound in air, the frequency \(f\) of the source is

\[f=\frac{v}{\lambda}=\frac{v}{4l}.\]

Alternatively, locate two successive resonance lengths \(l_1\) and \(l_2\). Then \(l_2-l_1=\lambda/2\), so that

\[f=\frac{v}{2(l_2-l_1)}.\]

(b)(ii)

For the fundamental note of a pipe closed at one end,

\[L=\frac{\lambda}{4}.\]

Thus,

\[\lambda=4L=4(1)=4\ \text{m}.\]

\[f=\frac{v}{\lambda}=\frac{340}{4}=85\ \text{Hz}.\]

Frequency of the note = \(85\ \text{Hz}\).

(c)