Question

A stationary source sends forth monochromatic sound with speed $v = 330$ m/s. A wall is moving towards it with speed $u = 33$ cm/s. How does the observed wavelength change after reflection from the wall?

• A. decreases by 0.2%
• B. increases by 0.2%
• C. decreases by 0.1%
• D. increases by 0.1%

Answer 1

Answer: (A)

decreases by 0.2%

Answer 2

The problem involves the Doppler effect for sound waves, occurring in two stages:

1. Sound waves from the stationary source incident on the moving wall.
2. Sound waves reflected from the wall (which acts as a moving source) back into the medium.

Let:

• $v$ be the speed of sound in the medium ($v = 330$ m/s).
• $u$ be the speed of the wall ($u = 33$ cm/s = $0.33$ m/s).
• $f$ be the frequency of the sound emitted by the stationary source.
• $\lambda$ be the wavelength of the sound emitted by the stationary source.

The original wavelength is $\lambda = \frac{v}{f}$.

Step 1: Frequency observed by the wall ($f'$).

The wall acts as an observer moving towards a stationary source. The formula for the observed frequency when an observer moves towards a stationary source is:

$f' = f \left( \frac{v + u}{v} \right)$

Step 2: Frequency of the sound reflected from the wall ($f''$).

The wall now acts as a source emitting sound of frequency $f'$. This source (the wall) is moving towards the original source's position (or effectively, towards a stationary observer in the medium). The formula for the observed frequency when a source moves towards a stationary observer is:

$f'' = f' \left( \frac{v}{v - u} \right)$

Substitute the expression for $f'$ into the equation for $f''$:

$f'' = f \left( \frac{v + u}{v} \right) \left( \frac{v}{v - u} \right)$

$f'' = f \left( \frac{v + u}{v - u} \right)$

Step 3: Wavelength of the reflected sound ($\lambda''$).

The wavelength of the reflected sound in the medium is given by $\lambda'' = \frac{v}{f''}$. Substitute the expression for $f''$:

$\lambda'' = \frac{v}{f \left( \frac{v + u}{v - u} \right)}$

$\lambda'' = \frac{v}{f} \left( \frac{v - u}{v + u} \right)$

Since $\lambda = \frac{v}{f}$, we have:

$\lambda'' = \lambda \left( \frac{v - u}{v + u} \right)$

Step 4: Percentage change in wavelength.

The change in wavelength is $\Delta \lambda = \lambda'' - \lambda$. The percentage change is $\frac{\Delta \lambda}{\lambda} \times 100\%$.

$\frac{\Delta \lambda}{\lambda} = \frac{\lambda'' - \lambda}{\lambda} = \frac{\lambda \left( \frac{v - u}{v + u} \right) - \lambda}{\lambda}$

$\frac{\Delta \lambda}{\lambda} = \left( \frac{v - u}{v + u} \right) - 1$

$\frac{\Delta \lambda}{\lambda} = \frac{(v - u) - (v + u)}{v + u}$

$\frac{\Delta \lambda}{\lambda} = \frac{v - u - v - u}{v + u}$

$\frac{\Delta \lambda}{\lambda} = \frac{-2u}{v + u}$

Now, substitute the given values:

$v = 330$ m/s $u = 0.33$ m/s

Percentage change $= \left( \frac{-2 \times 0.33}{330 + 0.33} \right) \times 100\%$

Percentage change $= \left( \frac{-0.66}{330.33} \right) \times 100\%$

Percentage change $\approx -0.00199806 \times 100\%$

Percentage change $\approx -0.1998\%$

Rounding to one decimal place, the percentage change is approximately $-0.2\%$. The negative sign indicates a decrease in wavelength.

Thus, the observed wavelength decreases by 0.2%.