IB Physics HL C Wave Behaviour Questions
Practise IB Physics HL wave behaviour through shared-core and HL resonance, interference, diffraction and standing-wave problems using measured data.
- Syllabus
- First assessment 2025
- Course
- Physics HL
- Level
- HL
Practise IB Physics HL wave behaviour through shared-core and HL resonance, interference, diffraction and standing-wave problems using measured data.
One end of a light spring is attached to a rigid horizontal support.
An object W of mass 0.15 kg is suspended from the other end of the spring. The extension x of the spring is proportional to the force F causing the extension. The force per unit extension of the spring k is 18Nm−1.
A student pulls W down such that the extension of the spring increases by 0.040 m . The student releases W and as a result W performs simple harmonic motion (SHM).
State what is meant by the expression "W performs SHM".
the acceleration of (force acting on) W is proportional to its displacement from equilibrium;
and directed towards equilibrium;
Determine the period of oscillation of the spring.
ω=xa;
=10.95 rad s−2;
T=(ω2π=)10.956.28=0.57 s;
W in (a) is immersed in a beaker of oil. As a result of this immersion the oscillations of W are critically damped. Describe what is meant by critically damped.
(the frictional force on W is such that) motion rapidly dies away/rapidly stops/stops in the minimum time;
without completing an oscillation / without overshooting (equilibrium position);
A spring, such as that in (a), is stretched horizontally and a longitudinal travelling wave is set up in the spring, travelling to the right.
Describe, in terms of the propagation of energy, what is meant by a longitudinal travelling wave.
The direction of oscillation of the particles of the medium is in the direction of energy propagation.
Marking guidance:
Must see "particles". Accept answer in terms of coils of spring in place of particles of medium.
The graph shows how the displacement x of one coil C of the spring varies with time t.
The speed of the wave is 3.0 cm s−1. Determine the wavelength of the wave.
frequency =(T1=0.801=)1.25 Hz;
wavelength =fv=1.253.0=2.4 cm or 2.4×10−2 m;
Draw, on the graph in (c)(ii), the displacement of a coil of the spring that is 1.8 cm away from C in the direction of travel of the wave, explaining your answer.
Graph: positive cosine;
line must cross the axis at 0.2 and 0.6 as shown.
Explanation: 1.8 cm is 3/4 of a wavelength.
Outline what is meant by a travelling wave.
The transfer/propagation of energy/momentum/information
Through oscillations/vibrations of medium/fields
Positions of maximum and minimum amplitude OR crests and troughs travel through a medium
A loudspeaker emits sound of frequency 210 Hz into a pipe with one open and one closed end. The diagram shows a representation of the standing wave established in the pipe.
The length of the pipe is 1.20 m .
Outline how the standing wave is formed in the pipe.
The incoming wave is reflected «from the closed end»
<<The reflected and incoming wave>> superpose/interfere
Determine the wavelength of the wave.
λ=⋖34L=34×1.20=>1.6<m≫
Calculate the speed of sound in the pipe stating the answer to an appropriate number of significant figures.
c=λf=1.60×210=336 OR 340 m s−1
Any answer to 2 OR 3 s.f.
Marking guidance:
Allow ECF from incorrect wavelength in bii)
The solid line represents the standing wave at time t and the dotted line represents the standing wave at an instant later. The dot is the equilibrium position of a particle P in the pipe. The up arrow indicates displacements to the right and the down arrow displacements to the left.
On the diagram, draw
a dot to indicate the approximate position of P at time t,
To the left of the equilibrium position on the same level
Accept any distance to the left
an arrow to indicate the velocity of P at time t.
Left horizontal arrow
Accept any arrow to the left inside the
tube.
The amplitude of oscillations of the standing wave in (b) is 4.2 mm . The mass of particle P in (c) is 1.8×10−6 kg.
Calculate
the total energy of P,
ω=≪2πf⇒2π×210=1.319×103≪rad s−1≫≪Energy=21mω2xo2=21×1.8×10−6×(1.319×103)2×(4.2×10−3)2=≫
MP1 can be awarded for a
correct substitution or value
[2]
2.7×10−5 OR 2.8×10−5≪ J≫
the displacement of P , when its kinetic energy is equal to its potential energy.
ALTERNATE 1
21mω2x2=21×(21mω2x02)x=2x0=24.2=3.0 «mm» ALTERNATE 221mω2(x02−x2)=21mω2x2x=2x0=24.2=3.0 «mm» ALTERNATE 3
Recognition that potential energy is 1.4×10−5≪ J>x=π1.8×10−6×(1.33×103)22×14×10−3=>3.0<mm≫<
For Alternate 3 allow ECF from the
energy value from di).
[2]
The frequency of sound is reduced to 140 Hz . Explain why a standing wave will not be formed in the pipe.
ALTERNATE 1
the pipe can only support standing waves with frequencies that are odd multiples of the first harmonic frequency
first harmonic frequency is 70 Hz
ALTERNATE 2
<<the new wavelength would be 2.4 m so>> a node would be formed at the open end
An antinode is required at the open end to form a standing wave
A converging lens is placed between an object and a screen. An image of the object is formed on the screen.
Draw a ray to locate the focal point of the lens. Label this point with the letter F .
a
one of the two rays above
The lens suffers from spherical aberration.
Draw lines to complete the rays in the diagram.
i
the extreme ray crosses principal axis closer than paraxial ray
Two loudspeakers, A and B, are driven in phase and with the same amplitude at a frequency of 850 Hz . Point P is located 22.5 m from A and 24.3 m from B. The speed of sound is 340 ms−1.
Deduce that a minimum intensity of sound is heard at P.
wavelength =850340=0.40 «m»
path difference =1.8 «m»
1.8 «m» =4.5λ OR 0.201.8=9 «half-wavelengths»
waves meet in antiphase «at P»
OR
destructive interference/superposition «at P»
Marking guidance:
Allow approach where path length is calculated in terms of number of wavelengths;
along path A (56.25) and path B (60.75) for MP2, hence path difference 4.5 wavelengths for MP3
A microphone moves along the line from P to Q . P Q is normal to the line midway between the loudspeakers.
B 咧
The intensity of sound is detected by the microphone. Predict the variation of detected intensity as the microphone moves from P to Q.
«equally spaced» maxima and minima
a maximum at Q
four «additional» maxima «between P and Q »
2 max
When both loudspeakers are operating, the intensity of sound recorded at Q is I0. Loudspeaker B is now disconnected. Loudspeaker A continues to emit sound with unchanged amplitude and frequency. The intensity of sound recorded at Q changes to IA.
Estimate I0IA.
the amplitude of sound at Q is halved
«intensity is proportional to amplitude squared hence» I0IA=41
In another experiment, loudspeaker A is stationary and emits sound with a frequency of 850 Hz . The microphone is moving directly away from the loudspeaker with a constant speed v. The frequency of sound recorded by the microphone is 845 Hz .
Explain why the frequency recorded by the microphone is lower than the frequency emitted by the loudspeaker.
speed of sound relative to the microphone is less
wavelength unchanged «so frequency is lower»
OR
fewer waves recorded in unit time/per second «so frequency is lower»
Calculate v.
845=850×340340−vv=2.00<m s−1>−