IB Physics HL C: Wave Behaviour

the acceleration of (force acting on) W is proportional to its displacement from equilibrium; and directed towards equilibrium;

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

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.

frequency $=\left(\frac{1}{T}=\frac{1}{0.80}=\right) 1.25 \mathrm{~Hz}$;
wavelength $=\frac{v}{f}=\frac{3.0}{1.25}=2.4 \mathrm{~cm}$ or $2.4 \times 10^{-2} \mathrm{~m}$;

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.

The incoming wave is reflected «from the closed end»
<<The reflected and incoming wave>> superpose/interfere

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$\lambda=\lessdot \frac{4 L}{3}=\frac{4 \times 1.20}{3}=>1.6<\mathrm{m} \gg$

[1]

$c=\ll \lambda f=1.60 \times 210 \Rightarrow \gg 336$ OR $340 \ll \mathrm{~m} \mathrm{~s}^{-1} \gg$
Any answer to 2 OR 3 s.f.

Marking guidance:

Allow ECF from incorrect wavelength in bii)

[2]

To the left of the equilibrium position on the same level

Accept any distance to the left

[1]

Left horizontal arrow

Accept any arrow to the left inside the
tube.

[1]

$$\begin{aligned} \omega=\ll 2 \pi f \Rightarrow 2 \pi \times 210=1.319 \times 10^{3} \ll \mathrm{rad} \mathrm{~s}^{-1} \gg \quad \ll \operatorname{Energy}=\frac{1}{2} m \omega^{2} x_{o}^{2}=\frac{1}{2} \times 1.8 \times 10^{-6} \times\left(1.319 \times 10^{3}\right)^{2} \times\left(4.2 \times 10^{-3}\right)^{2}=\gg \end{aligned}$$

MP1 can be awarded for a
correct substitution or value

[2]

$2.7 \times 10^{-5}$ OR $2.8 \times 10^{-5} \ll \mathrm{~J} \gg$

ALTERNATE 1
$$\begin{aligned} & \frac{1}{2} m \omega^{2} x^{2}=\frac{1}{2} \times\left(\frac{1}{2} m \omega^{2} x_{0}^{2}\right)^{} & x=\frac{x_{0}}{\sqrt{2}}=\frac{4.2}{\sqrt{2}}=3.0 \text { «mm» } & \text { ALTERNATE } 2 & \frac{1}{2} m \omega^{2}\left(x_{0}^{2}-x^{2}\right)=\frac{1}{2} m \omega^{2} x^{2} & x=\frac{x_{0}}{\sqrt{2}}=\frac{4.2}{\sqrt{2}}=3.0 \text { «mm» } & \text { ALTERNATE } 3 \end{aligned}$$

Recognition that potential energy is $1.4 \times 10^{-5} \ll \mathrm{~J}>$$$x=\pi \sqrt{\frac{2 \times 14 \times 10^{-3}}{1.8 \times 10^{-6} \times\left(1.33 \times 10^{3}\right)^{2}}}=>3.0<\mathrm{mm} \gg<$$

For Alternate 3 allow ECF from the
energy value from di).

[2]