C: SHM, waves, wave phenomena, standing waves and Doppler effect – IB Physics exam questions with mark scheme

1StandardHL only5 marks

A mass of 0.25 kg hangs on a spring of spring constant $40\ \mathrm{N\,m^{-1}}$ and oscillates vertically with simple harmonic motion of amplitude 6.0 cm.

(a) Calculate the time period of the oscillation. [2]

(b) Calculate the maximum speed of the mass. [1]

(d) Determine the total energy of the oscillation. [1]

$T=2\pi\sqrt{m/k}=2\pi\sqrt{0.25/40}$ seen (or $\omega=\sqrt{k/m}=\sqrt{40/0.25}=12.6\ \mathrm{rad\,s^{-1}}$)(1)
$T=0.50$ s (accept 0.497 s)(1)
$v_{max}=\omega A = 12.6\times0.060 = 0.76\ \mathrm{m\,s^{-1}}$ (0.76; allow 0.75 from rounded $T=0.50$ s; allow ECF from (a))(1)
$a_{max}=\omega^2A = 160\times0.060 = 9.6\ \mathrm{m\,s^{-2}}$ (9.6; allow 9.5 from rounded $T=0.50$ s; allow ECF)(1)
$E_T=\tfrac12 m\omega^2x_0^2 = \tfrac12\times0.25\times160\times0.060^2 = 0.072$ J (accept $\tfrac12 kx_0^2$ or $\tfrac12 mv_{max}^2$; allow 0.070–0.071 J from rounded values; allow ECF)(1)
Answer: (a) 0.50 s (b) 0.76 m s$^{-1}$ (c) 9.6 m s$^{-2}$ (d) 0.072 J
Examiner tip: Convert the amplitude from cm to m before substituting into $\omega A$ and $\omega^2 A$.

2HardHL only6 marks

The graph shows how the displacement $x$ of a $0.30\ \mathrm{kg}$ object undergoing simple harmonic motion varies with time $t$.

(a) Use the graph to determine the angular frequency of the motion. [2]

(b) Calculate the displacement of the object at $t=0.050\ \mathrm{s}$. [2]

Time period read from graph $T=0.40$ s (amplitude 8.0 cm)(1)
$\omega=2\pi/T = 15.7\ \mathrm{rad\,s^{-1}}$ (15.7; allow 16)(1)
Graph starts at maximum so $x=x_0\cos\omega t = 8.0\cos(15.7\times0.050)$ (calculator in radians)(1)
$x=5.7$ cm (5.7 cm; $5.7\times10^{-2}$ m; allow 5.6 cm if $\omega=16$ is used)(1)
$E_T=\tfrac12 m\omega^2x_0^2$ (amplitude in metres)(1)
$E_T=0.24$ J (0.24 J; allow 0.25 J if $\omega=16$ is used)(1)
Answer: (a) 15.7 rad s$^{-1}$ (b) 5.7 cm (c) 0.24 J
Examiner tip: Check that the calculator is in radian mode when evaluating $\cos\omega t$.

3Core4 marks

A wave travels along a rope. The graph shows the displacement $y$ of the rope against position $x$ at one instant. The frequency of the wave is $6.0\ \mathrm{Hz}$.

(a) Distinguish between transverse and longitudinal waves. [2]

(b) Use the graph to determine the speed of the wave. [2]

Transverse: oscillations (of particles) are perpendicular to the direction of energy transfer (accept direction of wave travel/propagation)(1)
Longitudinal: oscillations are parallel to the direction of energy transfer (accept direction of wave travel/propagation)(1)
Wavelength $\lambda=0.40$ m read from graph (one full cycle)(1)
$v=f\lambda = 6.0\times0.40 = 2.4\ \mathrm{m\,s^{-1}}$ (2.4)(1)
Answer: (a) transverse: oscillations perpendicular to the direction of energy transfer; longitudinal: oscillations parallel to it (b) 2.4 m s$^{-1}$
Examiner tip: This is a displacement–distance graph, so one full cycle along the $x$-axis is the wavelength, not the period; use $v=f\lambda$.

4Standard7 marks

A ray of light in air is incident on a glass block of refractive index 1.52 at an angle of incidence of $40^\circ$.

(a) Calculate the angle of refraction in the glass. [2]

(b) Calculate the speed of light in the glass. [1]

(d) State the two conditions required for total internal reflection to occur. [2]

$n_1\sin\theta_1=n_2\sin\theta_2$: $\sin\theta_2 = \sin40^\circ/1.52$(1)
$\theta_2=25.0^\circ$ (25°)(1)
$v=c/n = 3.00\times10^8/1.52 = 1.97\times10^{8}\ \mathrm{m\,s^{-1}}$ ($1.97\times10^8$)(1)
$\sin\theta_c = n_2/n_1 = 1/1.52$ (= 0.658)(1)
$\theta_c=41.1^\circ$ (accept 41°)(1)
Light travels from a medium of higher refractive index to one of lower refractive index (optically denser to less dense)(1)
Angle of incidence is greater than the critical angle(1)
Answer: (a) 25.0° (b) $1.97\times10^8$ m s$^{-1}$ (c) 41.1° (d) light travels from higher to lower refractive index; angle of incidence > critical angle
Examiner tip: Angles are measured from the normal, and the sine law has $n$ on the side of the matching angle.

5StandardHL only5 marks

Monochromatic light of wavelength $600\ \mathrm{nm}$ passes through a single slit of width $0.10\ \mathrm{mm}$. The diffraction pattern forms on a screen $2.0\ \mathrm{m}$ from the slit.

(a) Calculate the angular position of the first minimum. [2]

(b) Determine the width of the central maximum on the screen. [2]

$\theta=\lambda/b$ (small angle)(1)
$\theta=6.0\times10^{-3}$ rad(1)
Width $=2D\theta$ (central maximum spans both first minima)(1)
Width $=0.024$ m (2.4 cm; $2.4\times10^{-2}$ m)(1)
Width increases because $\theta\propto1/b$ (more diffraction through narrower slit)(1)
Answer: (a) $6.0\times10^{-3}$ rad (b) 0.024 m (c) wider
Examiner tip: The central maximum extends to the first minimum on each side, so double $D\theta$.

6Hard6 marks

In a double-slit experiment, light of wavelength $550\ \mathrm{nm}$ passes through two narrow slits $0.50\ \mathrm{mm}$ apart. The screen is $1.8\ \mathrm{m}$ from the slits.

(a) Calculate the fringe separation on the screen. [2]

(b) The whole apparatus is immersed in water of refractive index 1.33. Determine the new fringe separation. [2]

(c) In the original arrangement (in air), calculate the distance from the centre of the central maximum to the centre of the fourth bright fringe on one side of it. The central maximum is not counted as one of the four. [1]

(d) State, in the original arrangement, the path difference between the waves from the two slits at this fourth bright fringe. [1]

$s=\lambda D/d = 550\times10^{-9}\times1.8/0.50\times10^{-3}$(1)
$s=1.98$ mm (2.0 mm)(1)
Wavelength in water $\lambda'=\lambda/n=550/1.33=413.5$ nm (or $s'=s/n$)(1)
$s'=1.49$ mm (1.5 mm)(1)
$4s = 4\times1.98 = 7.9$ mm (7.92 mm; allow ECF from (a))(1)
$4\lambda=2.2\ \mu$m ($2.2\times10^{-6}$ m; accept “$4\lambda$”)(1)
Answer: (a) 2.0 mm (b) 1.5 mm (c) 7.9 mm (d) $2.2\ \mu$m
Examiner tip: Fringe separation is proportional to wavelength, and the wavelength shortens (frequency unchanged) in water.

7Standard5 marks

A pipe of length $0.60\ \mathrm{m}$ is closed at one end. The speed of sound in air is $340\ \mathrm{m\,s^{-1}}$.

(a) Calculate the frequency of the fundamental (first harmonic) standing wave in the pipe. [2]

(b) Calculate the frequency of the next harmonic that can be set up. [1]

(d) State the nature of the displacement at the closed end of the pipe. [1]

Fundamental has a quarter wavelength in the pipe: $\lambda=4L=2.4$ m(1)
$f=v/\lambda = 340/2.4 = 142$ Hz (accept 140 Hz)(1)
Next harmonic is the third: $3f_1 = 425$ Hz (accept 430 Hz)(1)
$f=v/2L = 340/1.2 = 283$ Hz (accept 280 Hz)(1)
A (displacement) node at the closed end (accept “zero displacement”; reject “pressure node”)(1)
Answer: (a) 142 Hz (b) 425 Hz (c) 283 Hz (d) node
Examiner tip: A closed pipe supports only odd harmonics; the second harmonic does not exist.

8StandardHL only6 marks

An ambulance siren emits sound of frequency $900\ \mathrm{Hz}$. The ambulance travels at a constant $25\ \mathrm{m\,s^{-1}}$ along a straight road, directly towards and then directly away from a stationary observer at the side of the road. The speed of sound is $340\ \mathrm{m\,s^{-1}}$.

(a) Calculate the frequency heard by the observer as the ambulance approaches. [2]

(b) Calculate the frequency heard as the ambulance moves away. [1]

(c) A spectral line from hydrogen has wavelength $656.3\ \mathrm{nm}$ in the laboratory but $659.6\ \mathrm{nm}$ in the light from a distant galaxy. Estimate the speed of the galaxy. Take $c=3.00\times10^8\ \mathrm{m\,s^{-1}}$. [2]

(d) State whether the galaxy is moving towards or away from the Earth. [1]

$f'=f\,v/(v-u_s) = 900\times340/(340-25)$(1)
$f'=971$ Hz (accept 970 Hz)(1)
$f'=900\times340/(340+25) = 838$ Hz (accept 840 Hz)(1)
$v\approx c\,\Delta\lambda/\lambda = 3.00\times10^8\times3.3/656.3$ (accept dividing by 659.6)(1)
$v=1.5\times10^{6}\ \mathrm{m\,s^{-1}}$ (accept $1.51\times10^6$)(1)
Away from the Earth (reason not required; e.g. wavelength increased / redshift)(1)
Answer: (a) 971 Hz (b) 838 Hz (c) $1.5\times10^6$ m s$^{-1}$ (d) away from Earth
Examiner tip: For a moving source use minus in the denominator when approaching and plus when receding.

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