Chapter 16, Problem 1Q

To determine

To rank:

The waves according to

a) their wave speed

b) the tension in the string along which they travel

Answer to Problem 1Q

Solution:

a) The waves can be ranked according to their wave speed as $v_1>v_4>v_2>v_3$(greatest first)

b) The waves can be ranked according to their tension in the string along which they travel as ${\tau }_1>{\tau }_4>{\tau }_2>{\tau }_3$(greatest first)

Explanation of Solution

1) Concept:

We can use the concept of the equation of transverse wave and speed of a travelling wave. The wave speed on a stretched string gives the relation between speed and tension in the string.

2) Formulae:

i) $y=y_m sin\left(kx-\omega t\right)$

ii)

$v=\frac{\omega }{k}$

iii)

$v=\sqrt{\frac{\tau }{\mu }}$

3) Given:

The four waves along the strings with the same linear densities are

i) $y_1=\left(3 mm\right) sin\left(x-3t\right)$

ii) $y_2=\left(6 mm\right) sin\left(2x-t\right)$

iii) $y_3=\left(1 mm\right) sin\left(4x-t\right)$

iv) $y_4=\left(2 mm\right) sin\left(x-2t\right)$

4) Calculations:

a) Rank the waves according to their wave speed :

The equation of transverse wave is

$y=y_m sin\left(kx-\omega t\right)\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

The speed of the travelling wave is

$v=\frac{\omega }{k}$

The equation (i),

$y_1=\left(3 mm\right) sin\left(x-3t\right)$

Compare this equation with equation (1), then the speed of the travelling wave is

$v_1=\frac{3}{1}$

$v_1=3$

The equation (ii) is

$y_2=\left(6 mm\right) sin\left(2x-t\right)$

Compare this equation with equation (1), then the speed of the travelling wave is

$v_2=\frac{1}{2}$

$v_2=0.5$

The equation (iii) is

$y_3=\left(1 mm\right) sin\left(4x-t\right)$

Compare this equation with equation (1), then the speed of the travelling wave is

$v_3=\frac{1}{4}$

$v_3=0.25$

The equation (iv) is

$y_4=\left(2 mm\right) sin\left(x-2t\right)$

Compare this equation with equation (1), then the speed of the travelling wave is

$v_4=\frac{2}{1}$

$v_4=2$

Hence, the rank of the waves according to the wave speed is $v_1>v_4>v_2>v_3$ (greatest fitst).

b) Rank the waves according to tension:

The wave speed on a stretched string is

$v=\sqrt{\frac{\tau }{\mu }}$

$v \alpha \sqrt{\tau }$

The speed on the stretched string is directly proportional to the tension in the string with the same linear density.

The speed on the stretched string for equation (i) is

$v_1 \alpha \sqrt{{\tau }_1}$

The speed on the stretched string for equation (ii) is

$v_2 \alpha \sqrt{{\tau }_2}$

The speed on the stretched string for equation (iii) is

$v_3 \alpha \sqrt{{\tau }_3}$

The speed on the stretched string for equation (i) is

$v_4 \alpha \sqrt{{\tau }_4}$

Hence, the rank of the waves according to their tension is ${\tau }_1>{\tau }_4>{\tau }_2>{\tau }_3$ (greatest first).

Conclusion:

We can find the wave speed by using its expression and rank their values. By using the expression of the speed on the stretched string, we can find thetension in each string and rank their values.