Oxford Revise AQA A Level Physics | Chapter P12 answers

P12: Simple harmonic motion

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Answers

Extra information

Mark

AO

Spec reference

01.1

Period = \(\frac{4.8\ \mathrm{s}}{3}\) = 1.6 s

\(f = \frac{1}{T} = \frac{1}{1.6\ \mathrm{s}}\) = 0.625 = 0.63 Hz

Evidence of use of graph to find T

Frequency

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3.6.1.1

01.2

\(
\begin{aligned}
\text{Maximum velocity } = \omega A & = 2\pi fA\\
& = 2 \times 3.14 \times 0.63 \times 0.02\\
& = 0.0785\ \mathrm{m\ s}^{-1} = 0.079\ \mathrm{m\ s}^{-1} = (7.9 \times 10^{-2}\ \mathrm{m\ s}^{-1})\\
\end{aligned}
\)

Evidence of use of frequency

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3.6.1.2

01.3

Find the maximum gradient

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3.4.1.3

01.4

Sinusoidal/same number of waves/frequency/periodic time

Inverted/a negative cosine graph

Maximum acceleration = \(\omega^2 A = \left( 2\pi f \right)^2 A\) = 0.308 m s–2 = 0.31 m s–2

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3.6.1.2

01.5

Condition for simple harmonic motion is that \(a \propto – x\)

So the graph of a is the same shape as that of x, but inverted

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3.6.1.2

02.1

Strategy:

States that readings of T (as the dependent variable) will be measured for different values of independent variable, wire diameter, d.

Clearly identifies at least 2 correct control variables: length/number of coils on spring/ mass

Make springs using wire of different diameters and measure the time period

Repeat measurements, omit outliers, find mean

Identifies dependent, independent and 2 control variables

Change d, measure T

Repeat, take mean

How to deal with outliers

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WS

02.2

Measure the time for 10 oscillations and divide the time by 10

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WS

02.3

Plausible reason, e.g. the length of wire is the same so the volume/mass of the wire will vary with the area of the wire, which is proportional to d 2

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3.4.2.1

02.4

Use the time period and mass to find k

\(T = 2\pi \sqrt{\frac{m}{k}}\)

 

\(k = \left( \frac{2\pi}{T} \right)^2\ \mathrm{m}\)

Plot a graph of k (y-axis) against d2 (x-axis), and if it is a straight line then the hypothesis is correct

Evidence of use of equation to find k

Correct axes identified

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3.6.1.3

03.1

\(
\begin{aligned}
& T = 2\pi \sqrt{\frac{m}{k}}\\
& \text{Plot a graph of }T\text{ against }\sqrt{\frac{1}{k}}\text{ : the gradient }=2\pi \sqrt{m}\\
\end{aligned}
\)

Or

\(\text{Plot }T^2\text{ against }\frac{1}{k}:\text{ gradient } = 4\pi^2m\)

Collect values of time period and spring constant

Change k, measure time period, use at least 6 different springs

Displace the trolley and measure the time for many oscillations with a stop clock, e.g. 5, and divide by 5 to find each time period

Repeat measurements and find the average time period for each value of k

Correctly identifies variables to plot, and how gradient relates to mass

Indication of range of independent variable

Accurate measurement of time

Repeat measurements

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3.6.1.3

03.2

Use the full reading on the stopwatch (to hundredths of a second) in measurements and calculation of the mean

Round up to one decimal place, and use uncertainty in using the stopwatch = \(\pm 0.2\ \mathrm{s}\) due to reaction time for both starting and stopping the stopwatch

Giving a total uncertainty of \(\pm 0.4\ \mathrm{s}\)

Use of full display on stopwatch until the calculation of final value

Estimation of reaction time

Total uncertainty is double the reaction time

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03.3

Suitable method:

Set up the light gate so that it is horizontal and triggered by the mass when it goes through its equilibrium position

Attach a straw/light rod to the mass that breaks the beam as the mass goes through its equilibrium position

The measurement of T will be double the time measured by the light gate

Suitable practical arrangement

Measurement of T that is accurate for the arrangement

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3.6.1.2

03.4

Each spring produces a restoring force of –kx, so the total restoring force = –2kx

ma = –2kx compared to ma = –kx

\(
\begin{aligned}
& \text{so } \omega^2 = \frac{2k}{m},\ \omega \text{ increases by }\sqrt2\\
& T = \frac{2\pi}{\omega}\text{ so } T\text{ is reduced by }\frac{1}{\sqrt2}\\
\end{aligned}
\)

Analysis to produce double the restoring force

Use of a = \(\omega\)2x

Answer

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3.6.1.2

04.1

For each length:

Allow the pendulum to swing 3 times (or more)

Take the times recorded by the light gate and double them to find the time period

Find the mean of all of the measurements

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3.6.1.3

04.2

x-axis length, y-axis T2

Line of best fit through (0, 0)

Line of best fit ignoring anomalous result, with gradient of \(\frac{4.0\ \mathrm{s}^2}{1\ \mathrm{m}}
\)

\(
\begin{aligned}
T & = 2\pi \sqrt{\frac{l}{g}}\\
T^2 & = 4\pi ^2 \frac{l}{g} \text{ so graph of } T^2 \text{ versus }l \text{ has a gradient of }\frac{4\pi^2}{g}\\
g & = \frac{4 \pi^2}{\text{ gradient }} = \frac{4 \pi^2}{4.0} = 9.9 (9.87)\ \mathrm{m\ s}^{-2}\\
\end{aligned}
\)

Both labels needed

Allow 3.9–4.1

Evidence of manipulation of equation

Allow 9.62–10.1

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3.6.1.3

04.3

Bigger – small angle approximation does not hold, bob may fall rather than swing, time period will be shorter than it should be

g will be smaller than it should be

Smaller – amplitude does not affect time period,

g not affected

Do not allow effect on g without explanation

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3.6.1.3

04.4

Systematic error in measurement of length

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3.4.2.2

05.1

The angle through which the pendulum is displaced should be small

so that you can use the small angle approximation

So that \(T = 2\pi \sqrt{\frac{l}{g}}\)/pendulum equation, which is independent of mass

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3.6.1.3

05.2

\(
\begin{aligned}
& x = A \cos\ \omega t\\
& A = 3.2 \times 10^{-2}\ \mathrm{m},\ \omega = \frac{2\pi}{T} = \frac{2\pi}{1.4} = 4.5 \text{ rad s}^{-1}\\
& x = 3.2 \times 10^{-2}\ \cos\ (4.5 t)\\
\end{aligned}
\)

Calculation of angular velocity

Equation

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3.6.1.2

05.3

\(
\begin{aligned}
& \text{Maximum velocity } = \omega A = 4.5 \times 0.032 = 0.14\ \mathrm{m\ s}^{-1}\\
& \text{Maximum kinetic energy } = \frac{1}{2}mv{^2} = \frac{1}{2} \times 0.26 \times \left( 0.14 \right)^2 = 2.7 \times 10^{-3}\ \mathrm{J}\\
\end{aligned}
\)

Graph that is correct shape (y = 1 –x2)

Maximum labelled, x -axis from –3.2 cm to + 3.2 cm

Calculation of maximum kinetic energy

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3.6.1.2

05.4

Assuming the total energy is constant,

the potential energy versus time graph is x2 graph

So that the kinetic energy + potential energy at any position = total energy

Or

\(\text{Total energy } = \frac{1}{2}\ kA^{2}\)

 

\(
\begin{aligned}
\text{So potential energy } & = \text{total energy} – \text{kinetic energy}\\
& = \frac{1}{2}\ k{A^2} – \frac{1}{2}\ m{v^2}\\
\end{aligned}
\)

Assumption

description

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3.6.1.3

05.5

The mass decreases, so kinetic energy decreases

The line will not be symmetrical/the line will reach a lower value

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3.6.1.3

06.1

Bathroom scales are compressed when you stand on them by an amount that is proportional to your weight/mass

In the International Space Station, both the scales and the astronaut are in free fall so the scales will not be compressed / gravitational field strength is lower

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3.4.1.1

06.2

\(
\begin{aligned}
T & = 2\pi \sqrt{\frac{m}{k}}\\
k & = m \left( \frac{2\pi}{T} \right)^2\\
& = 68.62 \times \left( \frac{2\pi}{2.084} \right)^2\\
& = 623.8 \text{ N m}^{-1}
\end{aligned}
\)

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3.6.1.3

06.3

0.9 × 68.62 kg = 61.76 kg

\(
\begin{aligned}
T & = 2\pi \sqrt{\frac{61.76\ \mathrm{kg}}{623.8\ \mathrm{N}\ \mathrm{m}^{-1}}}\\
& = 1.977\ \mathrm{s}\\
\end{aligned}
\)

(T is proportional to \(\sqrt{m}\) so as mass decreases so does periodic time)

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3.6.1.3

06.4

Max displacement = amplitude, which is proportional to energy

Energy transferred to thermal store due to friction

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3.6.1.3

06.5

No

The mass depends on the time period, which is independent of amplitude

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3.6.1.3

07.1

Volume of water displaced = A × x = 0.75 cm2 × 1.0 cm = 0.75 cm3

Mass of water = density of water × volume = 0.75 cm3 × 1 g cm–3

= 0.75 g = 7.5 × 10–4 kg

Weight = mg = 7.5 × 10–4 kg × 9.81 N kg–1 = 7.357 × 10–3 N

Correct use of equations for density and weight

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3.4.2.1

07.2

The restoring force is proportional to the distance that the tube is displaced from its equilibrium position OR F = −Agρx

Explanation of F ∝ x

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3.6.1.2

07.3

\(
\begin{aligned}
& \text{Acceleration } = \frac{F}{m} = \frac{7.4 \times 10^{-3}\,\mathrm{N}}{12 \times 10^{-3}\ \mathrm{kg}}\\
& a_{\text{max}} = 0.61\ \mathrm{m\ s}^{-2}\\
& a_{\text{max}} = \omega ^2A = \left( 2\pi f \right)^2 A\\
& f = \sqrt{\frac{a_{\max}}{A \left( 2\pi \right)^2}}\\
& f = \sqrt{\frac{0.61\ \mathrm{m\ s}^{-1}}{0.01\ \mathrm{m}\ \left( 2\pi \right)^2}}\\
& f = 1.2(4)\ \mathrm{Hz}\\
& T = \frac{1}{f} = \frac{1}{1.24\ \mathrm{Hz}} = 0.80\ \mathrm{s}\\
\end{aligned}
\)

Calculation of acceleration

Use of \(a_{\text{max}} = \omega^2 A\)

Alternatively, use \(a_{\text{max}} = \omega^2 A\) to find ω, then use \(T = \frac{2\pi}{\omega}\)

Answer

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3.6.1.2

07.4

Restoring force F = −Agρx

\(
\begin{aligned}
& a = – \frac{\text{area } \times \text{ g } \times \text{ density}}{\text{mass of tube}} \times x\\
& {\omega ^2} = – \frac{\text{area } \times g \times \text{ density}}{\text{mass of tube}} = \left( 2\pi f \right)^2 = \frac{2\pi^2}{T^2}\\
& \text{density }\propto \frac{1}{T^2}\\
& \text{A plot of density versus }\frac{1}{\left( \text{period} \right)^2}\text{ is a straight line}\\
\end{aligned}
\)

Derivation of value of \({\omega^2}\)

Manipulation to show time period

Answer

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3.4.2.1

07.5

A series circuit with an LDR and a fixed resistor

A cell/battery and a voltmeter across either the LDR or resistor

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3.5.1.5

08.1

\(k = \frac{F}{x} = \frac{700\ {\mathrm{N}}}{3.0 \times 10^{-2}\ {\mathrm{mm}}} = 23\ 000\ \mathrm{N\ m}^{-1}\)

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3.4.2.1

08.2

\(
\begin{aligned}
f & = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{\frac{23\,000}{1200}} = 0.70\ \mathrm{Hz}\\
& T = \frac{1}{f} = \frac{1}{0.70} = 1.4(2)\ \mathrm{s}\\
\end{aligned}
\)

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3.6.1.3

08.3

If the car goes over a bump/speed bump, it will displace the car from its equilibrium position

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3.6.1.2

08.4

\(
\begin{aligned}
& T = 2\pi \sqrt{\frac{m}{k}}\\
& \text{Either: plot } T^2 \text{ versus }m,\text{ gradient} = \frac{4 \pi^2}{k}\\
& \text{Or: plot } T \text{ versus }\sqrt{m}, \text{ gradient } = 2\pi \sqrt{\frac{1}{k}}\\
\end{aligned}
\)

Appropriate plot

Gradient that matches plot

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3.6.1.3

08.5

The oscillations are heavily/critically damped

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3.6.1.4

Skills box answers

Question

Answer

1

Use a fiducial marker (such as a pin) stuck at the equilibrium point of the mass.

Reduce parallax by observing oscillation at the same level as the fiducial marker/mass.

Use small displacements of the mass so that the mass hanger doesn’t ‘jump’ at the minimum displacement of the oscillation.

Include a measurement of reaction time in the measured time period.

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