6.0 Questions

Question1. Show that the equation, $y = a\sin (\omega t - kx)$ satisfies the wave equation $\frac{{{\partial ^2}y}}{{\partial {t^2}}} = {v^2}\frac{{{\partial ^2}y}}{{\partial {x^2}}}$. Find speed of wave and the direction in which it is travelling?

$$\frac{{{\delta ^2}y}}{{\delta {t^2}}} = - {\omega ^2}a\sin (\omega t - kx)$$

and $$\frac{{{\delta ^2}y}}{{\delta {x^2}}} = - {k^2}a\sin (\omega t - kx)$$

we can write these two equation as,

$$\frac{{{\delta ^2}y}}{{\delta {t^2}}} = \frac{{{\omega ^2}}}{{{k^2}}}.\frac{{{\delta ^2}y}}{{\delta {x^2}}}$$

comparing this with, $$\frac{{{\delta ^2}y}}{{\delta {t^2}}} = {v^2}.\frac{{{\delta ^2}y}}{{\delta {x^2}}}$$

we get, $$v = \frac{\omega }{k}$$

The neagtive sign between $\omega t$ and $kx$ implies that wave is travelling along positive $x$-direction.

Question 2. Equation of a transverse wave travelling in a rope is given by $y = 5\sin (4.0t - 0.02x)$, where $y$ and $x$ are expressed in cm and time in seconds. Calculate

(a) the amplitude, frequency, Velocity and wavelength of the wave.

(b) the maximum transverse speed and acceleration of a particle in the rope.

(a) Comparing this with the standard equation of wave motion

$$y = A\sin [2\pi ft - \frac{{2\pi }}{\lambda }x]$$

where $A$, $f$ and $\lambda $ are amplitude, fequency and wavelength, respectively.

Thus, amplitude $A$ = 5 cm

$$2\pi f = 4$$

$$ \Rightarrow Frequency,f = \frac{4}{{2\pi }} = 0.637cycles/s$$

Again, $\frac{{2\pi }}{\lambda }$ = 0.02 or Wavelength = $\frac{{2\pi }}{{0.02}}$ = 100$\pi $ cm

Velocity of the wave$v$ = $f\lambda $ = $\frac{4}{{2\pi }}\frac{{2\pi }}{{0.02}}$ = 200 cm/s

(b) Tranverse velocity of the particle, $u=\frac{{\delta y}}{{\delta t}}$ $$ = 5 \times 4\cos (4.0t - 0.02x)$$

$$ = 20\cos (4.0t - 0.02x)$$

Maximun velocity of the particle = 20 $cm/s$

Particle acceleration, $a$ = $\frac{{{\delta ^2}y}}{{\delta {t^2}}}$ $$ = - 20 \times 4\sin (4.0t - 0.02x)$$

Maximum acceleration of the particle = 80 $cm/{s^2}$

Question3. A uniform rope of mass 0.1 $kg$ and length 2.45 $m$ hangs from a ceiling.

(a) Find the speed of transverse wave in the rope at a point 0.5 $m$ distant from the lower end.

(b) Calculate the time taken by a transverse wave to travel the full length of the rope.

(a) As the string has mass and it is suspended vertically, tension in it will be different at different points. For a point at a distance $x$ from the free end, tension wiil be due to the weight of the string below it. So, if $m$ is the mass of string of length will be due to the weight of the string below it. So, if m is the mass of string of length $l$, the mass of length $x$ of the string will be,$\left( {\frac{m}{l}} \right)$

$$T = \left( {\frac{m}{l}} \right)xg = \mu xg$$

as$\left( {\frac{m}{l} = \mu } \right)$

$$\frac{T}{\mu } = xg$$

$$v = \sqrt {\frac{T}{\mu }} = \sqrt {xg} ..........(i)$$

At $x$ = 0.5 m, $$v = \sqrt {0.5 \times 9.8} = 2.21m/s$$

(b) From Eq. (i) the velocity of the wave is different at different points. So, if at point $x$ the wave travels a distance $dx$ in time $d$t, then

$$dt = \frac{{dx}}{v} = \frac{{dx}}{{\sqrt {gx} }}$$

$$\int_0^t {dt = \int_0^1 {\frac{{dx}}{{\sqrt {gx} }}} } $$

$$t = 2\sqrt {\frac{l}{g}} = 2\sqrt {\frac{{2.45}}{{9.8}}}=1.0s $$