IB Physics HL Theme C: Wave behaviour questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Question 1

Paper 2Easy6 marks

A position sensor tracks a component undergoing simple harmonic motion. Its displacement xx from equilibrium is modelled by

x=Acos⁡ϕ,ϕ=ϕ0+2πtT,x=A\cos\phi,\qquad \phi=\phi_0+\frac{2\pi t}{T},

where the phase angle ϕ\phi is in radians. The amplitude is 0.060 m0.060\ \mathrm{m} the period is 0.80 s0.80\ \mathrm{s} and the initial phase angle is π/3\pi/3

(a)

State the phase angle at t=0t=0

[1]
(b)

Calculate the phase angle at t=0.20 st=0.20\ \mathrm{s} in radians.

[2]
(c)

Calculate the displacement of the component at t=0.20 st=0.20\ \mathrm{s}

[2]
(d)

State the increase in phase angle during one complete oscillation, in radians.

[1]

Question 2

Paper 2Easy9 marks

A string on a musical instrument is fixed at both ends and vibrates in the mode shown in Figure 1.

00.10.20.30.40.50.6-2.5-2-1.5-1-0.500.511.522.5position, x / mdisplacement, y / mmAB
(a)

State the number of displacement nodes and the number of antinodes shown. Count the fixed ends as nodes.

[2]
(b)

The string has length 0.600 m0.600\ \mathrm{m} and the wave speed on it is 120 m s−1120\ \mathrm{m\,s^{-1}} For the mode shown, calculate the wavelength and frequency. Give both answers to 3 significant figures.

[3]
(c)

Compare the oscillation amplitudes at A and B, and state their phase difference.

[2]
(d)

Sound travels at 340 m s−1340\ \mathrm{m\,s^{-1}} in a pipe of length 0.600 m0.600\ \mathrm{m} The pipe is closed at one end and open at the other. Calculate the wavelength and frequency of the third harmonic. Ignore end correction.

[2]

Question 3

Multiple choiceEasy1 mark

A source emits light with wavelength 600 nm and moves directly away from an observer at 1.5×106 m s−11.5 \times 10^6\ \text{m s}^{-1} Take the speed of light as 3.0×108 m s−13.0 \times 10^8\ \text{m s}^{-1} What is the magnitude of the relative change in wavelength, expressed as a percentage?

Show answer and explanation

Answer: C. For light, the magnitude of the relative wavelength change is approximately v/cv/c Thus v/c=(1.5×106)/(3.0×108)=5.0×10−3=0.50%v/c=(1.5 \times 10^6)/(3.0 \times 10^8)=5.0 \times 10^{-3}=0.50\%

A: converts the fractional change to a percentage using a factor of 10 rather than 100.
B: introduces an erroneous factor of 1/21/2
D: shifts the decimal point one place too far when converting the fraction to a percentage.

Remember: multiply a fractional change by 100 to express it as a percentage.

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