Superposition Banker

1. (a) A stationary wave is set up on a string vibrating at 120 Hz. Adjacent nodes are 0.25 m apart. What is the speed of the progressive waves on the string?A 15 m s⁻¹B 30 m s⁻¹C 60 m s⁻¹D 120 m s⁻¹ [1]

1. (a) State the principle of superposition. [2]
2. (b) State two differences between a stationary wave and a progressive wave. [2]
3. (c) Explain how a stationary wave is formed on a stretched string fixed at both ends. [2]

A plastic pipe of length 0.68 m is open at both ends. Air in the pipe is made to vibrate. Speed of sound = 340 m s⁻¹. Ignore end corrections.

1. (a) State where the nodes and antinodes are when the pipe sounds its lowest note. [2]
2. (b) Calculate the frequency of the lowest note. [2]
3. (c) Calculate the frequency of the next higher note the pipe can produce. [1]

A string of length 1.20 m is fixed at both ends and made to vibrate by a vibration generator. When the frequency is 150 Hz a stationary wave with three loops (three antinodes) is formed.

1. (a) Calculate the wavelength of the waves on the string. [1]
2. (b) Calculate the speed of the waves on the string. [1]
3. (c) Determine the lowest frequency at which a stationary wave can be formed on this string, and the frequency that gives four loops. [2]
4. (d) State the number of nodes, including the ends, when there are four loops. [1]

A vertical tube, open at the top, has its lower end in water so that the length of the air column can be changed. A tuning fork of frequency 512 Hz is held over the top. Loud sounds are heard when the air column has lengths 0.165 m and 0.505 m (the first two resonances).

1. (a) Explain why a loud sound is heard at certain lengths of the air column. [2]
2. (b) Calculate the speed of sound in the tube. [2]
3. (c) The antinode at the open end is slightly outside the tube. Use the data to find how far outside the tube it is. [2]

1. (a) Two points P and Q on a stationary wave lie between the same pair of adjacent nodes. P is at the antinode and Q is close to one of the nodes. What is the phase difference between the oscillations of P and Q?A 0B 90°C 180°D it depends on the distance PQ [1]

A stationary wave of wavelength 0.60 m is set up on a long rope. One fixed end is at x = 0, which is a node. Points P, Q and R on the rope are at x = 0.15 m, x = 0.20 m and x = 0.45 m.

1. (a) State the positions of all the nodes between x = 0 and x = 0.90 m inclusive. [1]
2. (b) State the phase difference between the oscillations of (i) P and Q, (ii) P and R. [2]
3. (c) At one instant P is at its maximum upward displacement. Describe the displacements of Q and R at this instant. [2]
4. (d) State the displacement of every point on the rope a quarter of a period later. [1]

A string passes over a pulley and is kept under constant tension by a hanging mass. A vibration generator drives one end. The vibrating length L of the string is changed and the frequency f of the fundamental mode (one loop) is found each time:L = 0.400 m, f = 61 HzL = 0.500 m, f = 47 HzL = 0.600 m, f = 40 HzL = 0.700 m, f = 35 HzL = 0.800 m, f = 30 HzL = 0.900 m, f = 26 Hz

1. (a) Theory suggests f = v/(2L), where v is the speed of waves on the string. State the graph that should be plotted to give a straight line through the origin. [1]
2. (b) Calculate 1/L for each reading, with an appropriate unit. [1]
3. (c) The best-fit line through the origin has gradient 24.0 m s⁻¹. Determine the speed of the waves on the string. [1]
4. (d) Suggest two difficulties in deciding when the string is vibrating at its fundamental frequency, and an improvement for each. [4]

A microwave transmitter faces a flat metal plate. A detector is moved along the line between them. The distance between the first and the tenth minimum readings (9 intervals) is 13.5 cm. Speed of electromagnetic waves = 3.00 × 10⁸ m s⁻¹.

1. (a) Explain why the detector reading rises and falls as it is moved. [2]
2. (b) Calculate the wavelength and the frequency of the microwaves. [3]
3. (c) The readings at the minima are not zero. Suggest why, and explain why the minima are smaller close to the plate. [2]

A guitar string of vibrating length 0.65 m sounds its fundamental frequency of 196 Hz.

1. (a) Calculate the speed of the waves on the string. [2]
2. (b) The player presses the string against a fret so that the vibrating length is shorter and the fundamental frequency becomes 247 Hz. The tension does not change. Calculate the new vibrating length. [2]
3. (c) The string makes a sound wave in air of frequency 247 Hz. The speed of sound in air is 340 m s⁻¹. Calculate the wavelength of this sound and explain why it differs from the wavelength of the wave on the string. [2]

1. (a) A pipe closed at one end has a fundamental frequency of 120 Hz. Which of these frequencies can NOT be produced as a resonance of the same air column?A 240 HzB 360 HzC 600 HzD 840 Hz [1]

1. (a) Explain what is meant by coherent sources. [1]
2. (b) Explain why two separate light bulbs do not produce a visible interference pattern, whereas light from a single laser passing through two narrow slits does. [2]
3. (c) State one other condition needed for the dark fringes in a two-source pattern to be completely dark. [1]

1. (a) In a double-slit experiment with light, which single change increases the separation of the bright fringes on the screen?A moving the screen closer to the slitsB using slits that are further apartC using blue light instead of red lightD using slits that are closer together [1]

Two dippers S₁ and S₂ in a ripple tank vibrate in phase and produce water waves of wavelength 2.4 cm. Point P on the water surface is 15.6 cm from S₁ and 21.6 cm from S₂.

1. (a) Calculate the path difference at point P and state, with a reason, whether P is a point of maximum or minimum amplitude. [2]
2. (b) Point Q is 18.0 cm from S₁ and 22.8 cm from S₂. State whether Q is a maximum or a minimum. [1]
3. (c) The dippers are now adjusted so that they vibrate in antiphase. State what is now observed at P. [1]

Light from a laser passes through two narrow parallel slits 0.45 mm apart. An interference pattern is seen on a screen 2.40 m from the slits. The distance from the centre of the first bright fringe to the centre of the eleventh bright fringe is 32.0 mm.

1. (a) Calculate the fringe separation. [1]
2. (b) Calculate the wavelength of the light. [2]
3. (c) The light is replaced by light of wavelength 450 nm. Calculate the new fringe separation. [1]
4. (d) State and explain one reason why the width of 11 fringes is measured rather than the width of one. [1]

On a school sports ground, two loudspeakers 3.0 m apart are connected to the same signal generator and emit sound of frequency 680 Hz in phase. A student walks along a line parallel to the line joining the speakers and 12 m from it, and hears loud and quiet points. Speed of sound = 340 m s⁻¹.

1. (a) Calculate the wavelength of the sound. [1]
2. (b) State the path difference from the two loudspeakers to the first quiet point next to the central loud point. [1]
3. (c) Estimate the distance the student walks between adjacent loud points. [2]
4. (d) Suggest why the sound at the quiet points is not completely silent. [1]

1. (a) Two coherent sources emit waves of wavelength λ exactly in antiphase. At a point P the path difference from the two sources is 2λ. What is observed at P?A a maximum, because the path difference is a whole number of wavelengthsB a minimum, because the waves arrive in antiphaseC a maximum, because the waves arrive in phaseD neither, because the sources are not in phase [1]

A double slit with slit separation 0.30 mm is illuminated with white light. A screen is placed 1.5 m from the slits.

1. (a) Describe the pattern seen on the screen when white light is used. [3]
2. (b) Calculate the distance on the screen between the red (700 nm) and violet (400 nm) edges of the first-order bright fringe. [2]

A student shines a laser through a double slit. She records:width of 6 fringe spacings w = 14 ± 1 mmslit-to-screen distance D = 1.85 ± 0.01 mslit separation a = 0.50 ± 0.02 mm

1. (a) Calculate the wavelength of the laser light. [2]
2. (b) Calculate the percentage uncertainty in the wavelength. [2]
3. (c) Hence state the wavelength with its absolute uncertainty. [1]
4. (d) Suggest one change to the procedure that would most reduce the uncertainty. [1]

Microwaves from a single transmitter pass through two gaps in a metal screen. The centres of the gaps are 12.0 cm apart. A detector moved along a line parallel to the screen and 1.20 m from it finds the central maximum and then the first-order maximum 0.30 m to one side of it.

1. (a) Calculate the angle between the central line and the direction of the first-order maximum. [1]
2. (b) Use the condition for a maximum to calculate the wavelength of the microwaves. [2]
3. (c) Calculate the frequency of the microwaves. [1]
4. (d) A student uses x = λD/a instead. Calculate the wavelength this gives and explain why it is not reliable here. [2]

In a double-slit experiment the light waves from the two slits arrive at the screen with equal amplitude A, giving bright fringes and completely dark fringes. A filter is then placed over one slit so that the amplitude of the waves from that slit, at the screen, is halved.

1. (a) Describe and explain the change in the appearance of the bright and dark fringes. [3]
2. (b) Intensity is proportional to the square of the amplitude. Calculate the ratio of the intensity at a bright fringe to the intensity at a dark fringe after the change. [2]
3. (c) Calculate the intensity at a bright fringe after the change as a fraction of the intensity at a bright fringe before the change. [1]

Yellow light of wavelength 589 nm from a sodium lamp is incident normally on a diffraction grating with 300 lines per millimetre.

1. (a) Calculate the spacing between adjacent lines of the grating. [1]
2. (b) Calculate the angle of the first-order maximum. [2]
3. (c) Calculate the angle of the second-order maximum. [1]

Light of wavelength 600 nm is incident normally on a diffraction grating with 500 lines per millimetre.

1. (a) Determine the highest order of maximum that can be observed. [2]
2. (b) Calculate the angle of this highest-order maximum. [1]
3. (c) State the total number of bright maxima seen, including the zero order. [1]

A laser beam is directed normally at a diffraction grating with 400 lines per millimetre. On a screen 1.50 m from the grating, parallel to it, the two first-order spots are 0.82 m apart.

1. (a) Calculate the angle between the zero-order and first-order beams. [2]
2. (b) Calculate the wavelength of the laser light. [2]
3. (c) State the colour of the light, giving a reason. [1]

A student shines a green laser pointer of wavelength 532 nm normally at a diffraction grating whose label is missing. She measures the angle between the zero-order beam and the first-order beam as 18.6°.

1. (a) Calculate the number of lines per millimetre on the grating. [2]
2. (b) Suggest two reasons why a diffraction grating gives a more accurate value of wavelength than a double slit. [2]

Light from a hydrogen discharge tube contains a red line of wavelength 656 nm and a blue-green line of wavelength 486 nm. It falls normally on a grating with 600 lines per millimetre.

1. (a) Determine the highest order observed for each line. [2]
2. (b) Calculate the angle between the two lines in the second-order spectrum. [3]
3. (c) Explain why a grating with more lines per millimetre would separate the lines better, and state one disadvantage. [2]

1. (a) Monochromatic light falls normally on a diffraction grating. The second-order maximum is at 30.0° to the straight-through direction. At what angle is the first-order maximum?A 7.5°B 14.5°C 15.0°D 60.0° [1]

1. (a) State what is meant by diffraction. [1]
2. (b) Water waves of wavelength 1.5 cm pass through a gap in a barrier in a ripple tank. Describe how the pattern of waves beyond the gap changes as the gap width is reduced from 10 cm to 1.5 cm. [2]

1. (a) A person standing just outside an open doorway, out of sight, can be heard but not seen by someone inside the room. Which statement best explains this?A Sound waves are longitudinal but light waves are transverse.B Sound travels more slowly than light, so it has more time to bend.C The wavelength of sound is similar to the width of the doorway but the wavelength of light is much smaller.D Light is absorbed by the door frame but sound is reflected by it. [1]

A long-wave radio transmitter broadcasts at 198 kHz and an FM transmitter at 100 MHz. Speed of electromagnetic waves = 3.00 × 10⁸ m s⁻¹.

1. (a) Calculate the wavelength of each radio signal. [2]
2. (b) A village lies in a valley behind a range of hills about 1 km wide. Explain why the long-wave signal can be received in the village but the FM signal cannot. [2]