Waves for Cambridge International AS & A Level Physics

Reading cue. Count the loops between the supports, not the humps of the dashed curve: the dashed line is the same wave half a period later, and the 1.2 m spans all four loops.

1. aState State the number of nodes and the number of antinodes shown on the string in Fig. 4.1, counting the two fixed ends.

   recall2 marks

   Check answer · 2 marks

   1. 5 nodes
   2. 4 antinodes

2. bDetermine Determine the wavelength of the progressive waves on the string, and the distance along the string from a node to the nearest antinode.

   routine3 marks

   Check answer · 3 marks

   1. each loop is half a wavelength: loop length = 1.2 / 4 = 0.30 m
   2. λ = 0.60 m
   3. node to nearest antinode = λ/4 = 0.15 m

3. cExplain Explain why the three points between the supports stay at rest while the rest of the string moves between the solid and dashed positions.

   demanding3 marks

   Check answer · 3 marks

   1. two progressive waves of equal frequency and amplitude travel in opposite directions along the string (incident and reflected)
   2. at those points the two waves always arrive with a phase difference of 180°
   3. the displacements cancel at all times, so the resultant displacement there is permanently zero

4. dDeduce The pattern drawn in Fig. 4.1 is produced at 150 Hz. Deduce the next frequency above 150 Hz at which a clear stationary wave pattern appears on this string, and describe what is seen between the two frequencies.

   top of the paper4 marks

   Check answer · 4 marks

   1. four loops means the string is vibrating in its fourth mode, so the lowest possible frequency is 150 / 4 = 37.5 Hz
   2. the next pattern has five loops, at 5 × 37.5 Hz
   3. = 188 Hz
   4. between the two the string is not resonating: the loops lose their definition and the amplitude is small

Reading cue. θ is measured from the straight-through n = 0 beam, not from the grating face, and 500 lines per mm must first be inverted to give the line spacing d.

1. aCalculate Calculate the distance between the centres of adjacent lines on the grating labelled in Fig. 9.1.

   recall2 marks

   Check answer · 2 marks

   1. d = 1 mm / 500
   2. d = 2.0 × 10⁻⁶ m

2. bDetermine Determine the angle between the two first-order beams drawn either side of the straight-through direction.

   routine3 marks

   Check answer · 3 marks

   1. sin θ = λ/d = 590 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.295
   2. θ = 17.2°
   3. the two n = 1 beams are symmetrical about n = 0, so the angle between them is 34.3°

3. cDetermine The source is replaced by one giving white light of wavelengths from 400 nm to 700 nm. Determine the angular width of the first-order spectrum.

   demanding3 marks

   Check answer · 3 marks

   1. sin θ = 400 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.200, giving θ = 11.5°
   2. sin θ = 700 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.350, giving θ = 20.5°
   3. angular width = 20.5 − 11.5 = 9.0°

4. dDeduce With the white-light source still in place, deduce whether the second-order and third-order spectra overlap.

   top of the paper4 marks

   Check answer · 4 marks

   1. second order at 700 nm: sin θ = 2 × 700 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.700, θ = 44.4°
   2. third order at 400 nm: sin θ = 3 × 400 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.600, θ = 36.9°
   3. the third-order violet leaves at a smaller angle than the second-order red, so the two spectra overlap
   4. the third order is incomplete: sin θ would exceed 1 beyond about 670 nm, so its red end is missing