Studyledge

Waves: Question 1

Syllabus 7.1

Structured AS 7 marks

A line of buoys is anchored along a straight channel leading into a harbour. A water wave of frequency 0.40 Hz0.40\text{ Hz} and wavelength 3.5 m3.5\text{ m} travels along the channel at constant speed, passing each buoy in turn.

(a) State what is meant by the amplitude of a progressive wave. [1]

(b) State what is meant by the wavelength of a progressive wave. [1]

(c) Calculate the speed of the wave. [2]

(d) Calculate the period of the wave. [1]

(e) Two buoys, P and Q, lie along the direction of travel of the wave and are separated by a distance of 0.875 m0.875\text{ m}. Determine the phase difference between the oscillations of P and Q, giving your answer in degrees. [2]

Show worked solution Hide worked solution

Worked solution

Part (a): Amplitude of a progressive wave

The amplitude of a progressive wave is the maximum displacement of a point on the wave from its equilibrium (undisturbed) position.

Part (b): Wavelength of a progressive wave

The wavelength of a progressive wave is the distance between two adjacent points on the wave that are oscillating in phase, for example, the distance between two successive crests (or two successive troughs).

Part (c): Speed of the wave

The wave equation relates speed, frequency and wavelength: v=fλv = f\lambda

Substituting f=0.40 Hzf = 0.40\text{ Hz} and λ=3.5 m\lambda = 3.5\text{ m}: v=0.40×3.5=1.4 m s−1v = 0.40 \times 3.5 = 1.4\text{ m s}^{-1}

Check (rearranging the other way): 3.5×0.40=1.43.5 \times 0.40 = 1.4. Multiplication is commutative, so this confirms v=1.4 m s−1v = \boxed{1.4}\text{ m s}^{-1}.

Part (d): Period of the wave

The period is the reciprocal of the frequency: T=1f=10.40=2.5 sT = \frac{1}{f} = \frac{1}{0.40} = 2.5\text{ s}

Check: 2.5×0.40=1.02.5 \times 0.40 = 1.0, confirming that 2.5 s2.5\text{ s} and 0.40 Hz0.40\text{ Hz} are correctly reciprocal, so T=2.5 sT = \boxed{2.5}\text{ s}.

Part (e): Phase difference between P and Q

The separation of P and Q is a path difference of 0.875 m0.875\text{ m}. To convert this into a phase difference, express it as a fraction of one wavelength: path differenceλ=0.8753.5=0.25\frac{\text{path difference}}{\lambda} = \frac{0.875}{3.5} = 0.25

One full wavelength corresponds to a phase difference of 360°360° (a complete cycle), so: phase difference=0.25×360°=90°\text{phase difference} = 0.25 \times 360° = 90°

Check (independent method): 3.5 m×0.25=0.875 m3.5\text{ m} \times 0.25 = 0.875\text{ m}, confirming the fraction is exactly 14\tfrac14 of a wavelength, and 14×360°=90°\tfrac14 \times 360° = 90°, consistent with the first calculation. (Equivalently, 90°=π2 rad90° = \dfrac{\pi}{2}\text{ rad}.)

So the phase difference between P and Q is 90°\boxed{90}°.

Final answers

  • (a) Amplitude == maximum displacement from the equilibrium position
  • (b) Wavelength == distance between adjacent points in phase (e.g. successive crests)
  • (c) Speed =1.4 m s−1= \boxed{1.4}\text{ m s}^{-1}
  • (d) Period =2.5 s= \boxed{2.5}\text{ s}
  • (e) Phase difference =90°= \boxed{90}°