Waves: Question 6

Worked solution

Part (a): Frequency of a progressive wave

The frequency of a progressive wave is the number of complete oscillations (wave cycles) passing a fixed point per unit time (per second).

Part (b): Wavelength of the sound wave

Rearranging the wave equation $v = f\lambda$ to make $\lambda$ the subject: $\lambda = \frac{v}{f} = \frac{340}{250} = 1.36\text{ m}$

Check (rearranging the other way): $250 \times 1.36 = 340$, which matches the given speed, confirming $\lambda = \boxed{1.36}\text{ m}$.

Part (c): Distance travelled in 0.050 s

Distance travelled is speed multiplied by time: $\text{distance} = v \times t = 340 \times 0.050 = 17\text{ m}$

Check: $17 \div 0.050 = 340\text{ m s}^{-1}$, which reproduces the given speed, confirming the distance is $\boxed{17}\text{ m}$.

Part (d): Number of complete wavelengths in this distance

The number of complete wavelengths is the distance divided by the wavelength: $n = \frac{\text{distance}}{\lambda} = \frac{17}{1.36} = 12.5$

Check (independent method. Using $n = ft$ directly): the number of oscillations emitted in a time $t$ is also $n = f \times t = 250 \times 0.050 = 12.5$. This matches the value found by dividing distance by wavelength, confirming $n = \boxed{12.5}$.

Final answers

• (a) Frequency $=$ number of complete oscillations passing a fixed point per second
• (b) Wavelength $= \boxed{1.36}\text{ m}$
• (c) Distance $= \boxed{17}\text{ m}$
• (d) Number of wavelengths $= \boxed{12.5}$