Oxford Revise AQA A Level Physics | Chapter P4 answers

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P4: Progressive and stationary waves

Question

Answers

Extra information

Mark

AO

Spec reference

01.1

2.8 cm

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2

3.3.1.1

01.2

 \(
\begin{array}{l}
\lambda = 2.8\ \textrm{cm}\ c = 8.4\ \rm{cm} \rm{s}^{-1}\\
c = f\lambda \\
f = \frac{c}{\lambda } = \frac{{8.4}}{{2.8}} = 3\ {\rm{Hz}}\\
\end{array}
\)

Allow e.c.f for λ answer from 01.1

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2

3.3.1.1

01.3

 \(\frac{\pi }{2}
\textrm{ rad or } 90^{\circ}
\)

1 mark for phase difference,

1 mark for units

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3.3.1.1

01.4

 \(
\begin{array}{l}
\textrm{Displacement will be negative (downwards) to max in}\frac{T}{4}\ \rm{s}\\
\textrm{Decreases through to zero displacement at} \frac{T}{2}\ \rm{s}
\end{array}
\)

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3.3.1.1

02.1

Double-ended arrow/line from between two points in phase

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3.3.1.1

02.2

Maximum displacement

From the equilibrium position (from rest position/mean position)

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3.3.1.1

02.3

Longitudinal wave

Spring oscillates in a direction parallel to/in energy transfer

OR because of compressions and rarefactions

Rarefaction must be spelt correctly for second mark

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3.3.1.2

02.4

Stationary wave formed

By superposition (or interference) (of two progressive waves)

Stationary points are nodes

Destructive interference where the spring is stationary

Max of 3 marks awarded

max 3

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3.3.1.3

03.1

With unpolarised light, the vibrations are in many planes

In plane polarised light, the oscillations are in one plane only

ignore ‘direction’

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3.3.1.2

03.2

Reflected light is polarised

So intensity of light reflected on water will be reduced by polarising filter

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3.3.1.2

03.3

Rotate the polarising filter through 180°/360°

Variation in intensity between max and min (or light and dark)

One maxima and min in 180° OR two maxima (or two minima) in 360° rotation

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3.3.1.2

03.4

Sound waves are longitudinal waves

Since oscillations are parallel to/same direction as wave travel, they cannot be polarised

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3.3.1.2

03.5

Sketch:

  • showing ray of light reflected away from the normal as it leaves the water
  • straight line drawn showing where ray appears to come from person’s perspective

Explanation that change of speed at boundary causes refraction

Max 2 marks for two rays drawn and 1 mark for explanation

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3.3.2.3

04.1

Reflection from metal plate

Two waves of the same frequency/wavelength

travelling in opposite directions (or forward/reflected waves)

Maxima where waves are in phase or interfere constructively

Minima where waves are out of phase/antiphase or interfere destructively

Nodes and antinodes or stationary waves identified

Any three awarded

max 3

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3.3.1.3

04.2

 \(
\begin{array}{l}
\textrm{Distance between minima is}\ \frac{\lambda }{2}\\
4 \times \frac{\lambda }{2}= 54\ \rm{mm}\\
\lambda = 27\ \rm{mm}
\end{array}
\)

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3.3.1.3

3.3.1.1

04.3

\(
\begin{array}{l}
c =f\ \lambda\ \textrm{and}\ c = 3.0 \times 10^8\ \rm{m} \rm{s}^{-1}\\
f = \frac{c}{\lambda } = \frac{{3.0 \times {{10}^8}}}{{27 \times {{10}^{-3}}}} = 1.1 \times 10^{10}\ \rm{Hz}\\
11\ \rm{GHz}
\end{array}
\)

Allow e.c.f from 04.2

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3.3.1.3

3.3.1.1

3.1.1

04.4

P labelled close to the plate in direct line with transmitter

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3.3.1.3

04.5

The distance travelled by the transmitted wave and the reflected wave is similar at point P

The amplitude of both waves will be similar

Max destructive interference

Max of two points

max 2

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3.3.1.3

04.6

The microwave transmitter produces plane polarised waves and so the detector must be in the correct plane

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3.3.1.2

05.1

Stationary waves set up in the microwave

Microwaves are reflected in the oven

Two waves of the same frequency/wavelength travelling in opposite directions (or forward/reflected waves)

Melted marshmallows are where waves are in phase or interfere constructively

The melted marshmallows are at an antinode

Any three from

max 3

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3.3.1.3

05.2

 \(
\begin{array}{l}
c = f\ \lambda\ \textrm{and}\ c = 3.0 \times 10^8\ \textrm{m s}^{-1}\\
\lambda = \frac{c}{f} = \frac{{3.0 \times {{10}^8}}}{{2450 \times {{10}^6}}} = 0.122\ \rm{m}\\
\textrm{Distance between minima is}\ \frac{\lambda }{2}\\
= 0.061\ \rm{m}
\end{array}
\)

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3.3.1.1

3.3.1.3

3.1.1

05.3

 \(
\begin{array}{l}
c = f \lambda = 2450 \times 10^6 \times 0.142\ \rm{m} = 3.48 \times 10^8\ \textrm{m s}^{-1}\\
\%\ {\rm{error}} = \frac{{{\rm{difference}}}}{{{\rm{actual}}}} \times 100\%\\
\%\ {\rm{error}} = \frac{{{4}{.8} \times {1}{{0}^7}}}{{{3}{.0} \times {1}{{0}^8}}}
\times 100\% = 16\%
\end{array}
\)

Allow e.c.f from value of c

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3.3.1.3

3.1.2

05.4

They would have to remove/disable the turntable so that marshmallows were stationary

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05.5

In a progressive wave, all particles vibrate at same frequency

AND in a stationary wave, all particles except those at nodes vibrate at same frequency

In a progressive wave, all particles have same amplitude

AND in stationary waves, particles vary in amplitude from zero at nodes to max at antinodes

In progressive waves, the phase difference between particles \(=
\frac{{2\pi d}}{\lambda }\), where d is the distance apart

AND in stationary waves, phase difference = nπ, where n is the number of nodes between the particles

Each statement must compare stationary and progressive waves

2 marks maximum for each property compared

Two properties compared for maximum marks

max 4

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3.3.1

06.1

Waves travel to the boundaries and are reflected

Two waves travelling in opposite directions interfere/superpose

Antinodes or maxima where waves are in phase or interfere constructively

OR

Nodes/minima where waves are out of phase/antiphase or interfere destructively

Nodes form at fixed ends

NOT bounces off

max 3

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3.3.1.3

06.2

λ = 0.62 × 2 = 1.24 m

c = f λ = 82.41 × 1.24 = 102 m s−1

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3.3.1.1

06.3

 \(f = \frac{c}{\lambda } = \frac{102}{{0.62}}= 165\ \rm{Hz}\)

Allow e.c.f from 06.2

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3.3.1.1

06.4

 \(
\begin{array}{l}
\textrm{Use of}\ f = \frac{1}{{2l}}\sqrt {\frac{T}{\mu }} \\
f{^2}\ \propto \frac{1}{\mu }\\
{f^{2}}\,\mu = \textrm{constant} = 82.41^{2} \times 6.44 \times 10^{-3}\\
\rm{New}\ \mu = \frac{{{{82.41}^2} \times 6.44 \times {{10}^{-3}}}}{{{{196}^2}}} = 1.14 \times 10^{-3}\ \rm{kg}\ \rm{m}^{-1}
\end{array}
\)

Alternatively, they could work out T from initial info and calculate

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3.3.1.3

07.1

3rd/third harmonic

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3.3.1.3

07.2

P antinode AND constructive interference/in phase

Q node AND destructive interference/out of phase

Need name and description for each mark

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3.3.1.3

07.3

P has positive amplitude and Q negative

P and Q are π rad or 180° out of phase

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3.3.1.3

3.3.1.1

07.4

 \(
\begin{array}{l}
\lambda = \frac{2}{3}L\ = 0.80\ \rm{m}\\
c = f \lambda\\
f = \frac{c}{\lambda } = \frac{{13.6}}{{0.80}} = 17\ \rm{Hz}
\end{array}
\)

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3.3.1.1

3.3.1.3

07.5

6th harmonic sketched – expect to see 6 antinodes and 7 nodes

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3.3.1.3

08.1

 \(
\begin{array}{l}
80\ \textrm{(ms)}\\
f = \frac{1}{T}= 12.5 Hz\\
f^2 = 156\ \rm{Hz}
\end{array}
\)

Answer in table row should be completed

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3.3.1.1

08.2

\(
\begin{array}{l}
T = mg\\
f = \frac{1}{2l} \sqrt {\frac{T}{\mu }} =\frac{1}{{2l}}\sqrt {\frac{{mg}}{\mu }}\\
f{^2}\ = \frac{1}{{4{l^2}}} \times \frac{{mg}}{\mu }\\
\textrm{Since}\ l,\ g,\ \textrm{and}\ \mu\textrm{ are constant}\\
{f^2} \propto m
\end{array}
\)

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3.3.1.3

08.3

 \(\textrm{Mark for plotting point} \pm \frac{1}{2} \textrm{square on graph}\)

Mark for drawing line of best fit

Large triangle drawn or evidence shown

1.2 ± 0.1 (g Hz−2 )

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MS 3.2, 3.3, 3.4

08.4

 \(
\begin{array}{l}
\textrm{Gradient} =\frac{g}{{4{l^2}\mu }}\\
\mu = \frac{{9.81}}{{4 \times {1^{{\kern 1pt} 2}} \times 1.2}}\\
2.0\ \rm{g}\ \rm{m}^{-1}
\end{array}
\)

e.c.f

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3.3.1.3

08.5

 \(
\begin{array}{l}
\textrm{% uncertainty in length} =
\frac{{0.001}}{{1.000}} \times 100\% = 0.1\%\\
\textrm{OR % uncertainty in mass}=\frac{{0.1}}{{1.7}} \times 100\% = 5.9\%\\
\textrm{total % error = 6.0%}\\
\textrm{absolute error} = 0.06 \times 1.7 = \pm\ 0.1\ \rm{g}\ \rm{m}^{-1}
\end{array}
\)

1 mark for calculating either % uncertainty

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3.1.2

08.6

 \(
\begin{array}{l}
\% \ {\rm{difference}} = \frac{{{\rm{difference}}}}{{{\rm{actual}}}}\times 100\%\\
\% \ {\rm{difference}} = \frac{{{0}{.3}}}{{{1}{.7}}} \times\ 100\% = 18\%
\end{array}
\)

Possible e.c.f from their value for 08.4

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3.1.2

Skills box answers

Question

Answer

1

Wear goggles in case the wire snaps. Place a box containing padding material below the masses in case they fall.

2

Set up the apparatus as shown in the diagram. Record values of f for different values of l. Change the length in the string by moving the vibrator closer to the pulley. \(f = \frac{1}{2l}\sqrt {\frac{T}{\mu }} \), therefore plot a graph of l against \(\sqrt{T},\) will produce a straight line through the origin, confirming the relationship between l and T.

3

 \(
\begin{array}{l}
\textrm{The gradient of the graph }= \frac{1}{{\left( {4{l^2}\mu } \right)}}\\
\textrm{so } \mu = (4l^2 \textrm{ gradient})^{-1}
\end{array}
\)
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