First slide

Reflection and refraction of waves

Question

A long composite string is made up by joining two strings of different mass per unit length $μ and 4 μ$ respectively. The composite string is under the same tension. A transverse wave pulse $Y = (6 mm) \sin (5 t + 40 x),$ where ‘t’ is in seconds and ‘x’ is in metres, is sent along the lighter string towards the joint. The joint is at $x = 0$ . The equation of the wave pulse reflected from the joint is

Moderate

A

$Y = (2 mm) \sin (5 t - 40 x)$
B

$Y = (4 mm) \sin (40 x - 5 t)$
C

$Y = (- 2 mm) \sin (5 t - 40 x)$
D

$Y = (2 mm) \sin (5 t - 10 x)$

Solution

Given wave pulse $Y = (6 mm) \sin (5 t + 40 x) ...... (1)$
We know speed of wave $v = \sqrt{\frac{T}{μ}} .... (2)$
Assume $v_{1} =$ Speed in string 1 (of density $μ_{1}$ )
$v_{2} =$ Speed in string 2 (of density $μ_{2}$ )
$v_{1} = \sqrt{\frac{T}{μ}}; v_{2} = \sqrt{\frac{T}{4 μ}}$
$\Rightarrow \frac{v_{1}}{v_{2}} = 2$
$∴ v_{2} < v_{1}$
$\Rightarrow$ 2nd medium is denser
$∴$ The wave reflected from the denser medium has phase change of Amplitude of reflected wave is
$\Rightarrow A_{r} = \frac{v_{2} - v_{1}}{v_{2} + v_{1}} \times A_{i}$
Amplitude of incident wave
Substituting the values we get
$\Rightarrow A_{r} = \frac{v_{2} - v_{1}}{v_{2} + v_{1}} \times 6 = \frac{\frac{v_{1}}{2} - v_{1}}{\frac{v_{2}}{2} + v_{1}} \times 6 = - 2 mm$
$∴ equation of reflected wave pulse is Y = (- 2 mm) \sin (5 t - 40 x)$
Therefore, the correct answer is (C).

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