EduNinja

[Maximum number: 5]

(a)

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 $18 \mathrm{Nm}^{-1}$ .

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).

[ 5 ]

(i)

State what is meant by the expression "W performs SHM".

[ 2 ]

(ii)

Determine the period of oscillation of the spring.

[ 3 ]

$\omega=\sqrt{\frac{a}{x}}$ ;
$=10.95 \mathrm{~rad} \mathrm{~s}^{-2}$ ;
$T=\left(\frac{2\pi}{\omega}=\right) \frac{6.28}{10.95}=0.57 \mathrm{~s}$ ;

[Maximum number: 15]

This question is in two parts. Part 1 is about simple harmonic motion (SHM). Part 2 is about gas in an engine.

Part 1 Simple harmonic motion (SHM)
An object is placed on a frictionless surface. The object is attached by a spring fixed at one end and oscillates at the end of the spring with simple harmonic motion (SHM).

The tension F in the spring is given by F=k x where x is the extension of the spring and k is a constant.

(a)

Show that $\omega^{2}=\frac{k}{m}$ .

[ 2 ]

m a=-k x;
$a=-\frac{k}{m} x ;($ condone lack of negative sign )

\left(\omega^{2}=\frac{k}{m}\right)

or
implied use of defining equation for simple harmonic motion $a=-\omega^{2} x$ ;
(so $\omega^{2}=\frac{k}{m}$ )
m a=-k x so $a=-\left(\frac{k}{m}\right) x$ ;

(b)

One cycle of the variation of displacement with time is shown for two separate mass-spring systems, A and B .

[ 5 ]

(i)

Calculate the frequency of the oscillation of A.

[ 1 ]

$0.833(\mathrm{~Hz})$ ;

(ii)

The springs used in A and B are identical. Show that the mass in A is equal to the mass in B.

[ 2 ]

frequency/period is the same so $\omega$ is the same;
k is the same (as springs are identical);
(so m is the same)

(iii)

Outline how you would use the graph to confirm that A is performing simple harmonic motion.

[ 2 ]

defines simple harmonic motion as acceleration proportional to negative displacement $/ x=x_{0} \sin \omega t$ ;
correct method for showing how graph leads to the definition;

(c)

The graph shows the variation of the potential energy of A with displacement.

On the axes,

[ 5 ]

(i)

draw a graph to show the variation of kinetic energy with displacement for the mass in A. Label this A.

[ 2 ]

(ii)

sketch a graph to show the variation of kinetic energy with displacement for the mass in B. Label this B.

[ 3 ]

end displacements correct $\pm 0.01 \mathrm{~m}$ ;
maximum lower than 0.16 J ;
maximum equal to $0.04 \mathrm{~J} \pm$ half square;

(d)

Using data from (b) and (c), calculate the mass in A.

[ 3 ]

maximum speed of oscillator $\left.\left(=\omega x_{0}=2 \pi \times 0.833 \times 0.2\right)=1.05\left(\mathrm{~m} \mathrm{~s}^{-1}\right) ;\right\} \begin{aligned} & (\boldsymbol{N} . \boldsymbol{B} . \\ & \left.\omega=5.24 \mathrm{rad} \mathrm{s}^{-1}\right)\end{aligned}$
maximum energy read-off $=0.16(\mathrm{~J}) ;\} \begin{aligned} & \text { (condone ECF for incorrectly drawn } \\ & \text { maximum kinetic energy value in (c)(i)) }\end{aligned}$
use to $\frac{1}{2} m v^{2}$ to give mass value ( $=0.292 \mathrm{~kg}$ using correct data);
or
read-off from graph of $\mathrm{ke}_{\text {max }}=0.16(\mathrm{~J})$ and $x_{\text {max }}=0.2(\mathrm{~m})$ ; (both needed)
$0.16=\frac{1}{2} m \times(2 \pi f)^{2} \times(0.2)^{2} ;$
0.292 (kg); (use of 0.8 Hz gives 0.32)

Allow any alternative method if correct, eg: T from graph, k evaluated $\left(8 \mathrm{Nm}^{-1}\right)$ .

Part 2 Gas in an engine

[Maximum number: 9]

This question is in two parts.

(a)

A particle P moves with simple harmonic motion.

[ 2 ]

(i)

State, with reference to the motion of P , what is meant by simple harmonic motion.

[ 2 ]

(b)

The graph shows how the velocity v of particle P varies with time t.

Use the graph opposite to determine for the motion of P the

[ 7 ]

(i)

period.

[ 1 ]

(ii)

amplitude.

[ 4 ]

Marking guidance:

max velocity $=0.74( \pm 0.02) \mathrm{ms}^{-1}$ ;
recognize max velocity $=\omega x_{0}$ ;
$\omega=\left(\frac{2 \pi}{T}=\frac{2 \pi}{0.30}=\right) 20.9 \mathrm{rads}^{-1} ;$ $x_{0}=\left(\frac{0.74}{20.9}=\right) 3.5( \pm 0.2) \times 10^{-2} \mathrm{~m} ;$
or
identifies displacement with area; uses one quarter of a cycle;
answer in the range of 30 to 40 mm ;
answer in the range of 33 to 37 mm ;

(iii)

displacement of P from equilibrium at $t=0.2 \mathrm{~s}$ .

[ 2 ]

$\quad v=0.64( \pm 0.2) \mathrm{ms}^{-1}$ ;
use $v=\omega \sqrt{\left(x_{0}^{2}-x^{2}\right)}$ to get $x=1.7( \pm 0.2) \times 10^{-2} \mathrm{~m}$ ;
or
recognition that $x=x_{0} \cos \omega t$ ;
$x\left(=35 \cos \left[\frac{2 \pi}{0.3} \times 0.2\right]\right)=17.5 \mathrm{~mm} ;$

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IB Physics HL C 1 Simple Harmonic Motion Topic Practice

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IB Physics HL C 1 Simple Harmonic Motion Topic Practice

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