IB Physics SL Theme C: Wave behaviour questions

Answer: D. For a mass–spring system, $T=2\pi\sqrt{m/k}$ and $\omega=2\pi/T$

A: gives the period ratio but does not take its reciprocal for angular frequency.
B: applies the mass change alone, giving a factor of $2$ and omits the spring change.
C: applies the spring change alone, giving a factor of $3$ and omits the mass change.
Remember: combine both changes when finding the period, then use the inverse relation between angular frequency and period.

Answer: C. The graph gives particle displacement along $y$ while the wave travels along $x$ therefore particle motion is perpendicular to the direction of travel. A wave that displaces particles of a cord is mechanical and requires a material medium, so it cannot propagate through a vacuum.

A: reads the horizontal position axis as the particle-motion direction and assumes the wave can travel without a medium.
B: reads the horizontal position axis as the particle-motion direction, giving parallel motion.
D: identifies the perpendicular motion correctly but mistakes a mechanical wave for one that can propagate through a vacuum.

Remember: mechanical waves require a medium; electromagnetic waves do not.

Answer: D. Each ray is perpendicular to its wavefronts, so the angles to the normal are $\alpha$ and $\beta$ Frequency is unchanged at the boundary, so $\lambda_2/\lambda_1=v_2/v_1$ From Snell’s law, $n_1\sin\alpha=n_2\sin\beta$ and $n=c/v$ giving $v_2/v_1=n_1/n_2=\sin\beta/\sin\alpha$
A: reverses the incident-to-transmitted ratio.
B: uses the angle ratio instead of the sine ratio.
C: treats the marked angles as angles between rays and the boundary, rather than between wavefronts and the boundary.
Remember: the angle between a wavefront and the boundary equals the angle between its ray and the normal.

Answer: B. The source and observer move in the same direction, but the source is faster, so it approaches the observer at a relative speed of $0.00200c$ The wavefronts are encountered more frequently, giving a fractional frequency increase of $0.00200$ Thus $f_{\text{obs}}=(5.000\times10^{14})(1+0.00200)=5.010\times10^{14}\,\text{Hz}$

A: Uses the observer’s speed alone, predicting a $0.00100$ decrease in frequency.
C: Uses the source’s speed alone and ignores the observer’s motion.
D: Adds the speeds despite their being in the same direction, giving a fractional increase of $0.00400$

Remember: for motion in the same direction, the relative speed is the difference between the speeds.