Waves in Air, Fluids and Solids

Solution

1. 1

   Identify the direction the wave is travelling.

   $wave travels horizontally along the rope$

2. 2

   Identify the direction the rope's particles move.

   $particles move vertically (up and down)$

3. 3

   Compare the two directions and conclude.

   $particle motion is perpendicular to wave motion \Rightarrow \text{transverse}$

Solution

1. 1

   Describe what a sound source (e.g. a loudspeaker) does to the air.

   $\text{cone pushes air forwards then back, repeatedly}$

2. 2

   State what the air molecules do.

   $\text{air molecules oscillate parallel to the wave direction}$

3. 3

   Conclude using the AQA definition.

   $\text{parallel motion} \Rightarrow \text{longitudinal}$

Solution

1. 1

   State what sound needs.

   $\text{sound is longitudinal and requires a medium of particles}$

2. 2

   Note conditions on the Moon's surface.

   $\text{Moon has effectively no atmosphere}$

3. 3

   Reach the conclusion and contrast with EM.

   $\text{no air} \Rightarrow \text{no sound; radio (EM) travels through vacuum}$

Solution

1. 1

   Recall T = 1/f.

   $T = 1/f$

2. 2

   Substitute f = 50 Hz.

   $T = 1/50 = 0.02\,\text{s}$

Solution

1. 1

   Convert wavelength to metres.

   $\lambda = 4.0\,\text{cm} = 0.040\,\text{m}$

2. 2

   Apply the wave equation.

   $v = f\lambda = 12 \times 0.040$

3. 3

   Calculate.

   $v = 0.48\,\text{m/s}$

Solution

1. 1

   Convert frequency to Hz.

   $f = 93.5\,\text{MHz} = 93.5 \times 10^6\,\text{Hz}$

2. 2

   Rearrange v = fλ for λ.

   $\lambda = v/f$

3. 3

   Substitute and evaluate.

   $\lambda = \dfrac{3 \times 10^8}{93.5 \times 10^6} \approx 3.21\,\text{m}$

Solution

1. 1

   Convert period to seconds.

   $T = 2.5\,\text{ms} = 0.0025\,\text{s}$

2. 2

   Use f = 1/T.

   $f = 1/0.0025 = 400\,\text{Hz}$

Solution

1. 1

   Each loop is half a wavelength. Three loops span 0.60 m so 1.5λ = 0.60 m.

   $\lambda = 0.60 / 1.5 = 0.40\,\text{m}$

2. 2

   Apply the wave equation.

   $v = f\lambda = 50 \times 0.40 = 20\,\text{m/s}$

Solution

1. 1

   The angle from the surface is 35°, so the angle from the normal is 90° − 35° = 55°.

   $i = 90 - 35 = 55^\circ$

2. 2

   Apply the law of reflection.

   $r = i = 55^\circ$

Solution

1. 1

   After reflection from mirror 1, angle of reflection is 30° from the normal of mirror 1.

   $r_1 = 30^\circ$

2. 2

   The two mirrors are at 90°, so the normals are also at 90°. By geometry, the ray strikes mirror 2 at (90° − 30°) = 60° from the second normal.

   $i_2 = 90 - 30 = 60^\circ$

Solution

1. 1

   Describe the surface.

   $\text{paper is rough on a microscopic scale}$

2. 2

   Apply the law of reflection at each micro-mirror.

   $\text{each tiny part still obeys } i = r$

3. 3

   Add up over many tilts → diffuse reflection.

   $\text{rays scatter in all directions} \Rightarrow \text{diffuse reflection}$

Solution

1. 1

   The ideal gradient is exactly 1 if i = r.

   $\text{ideal gradient} = 1.00$

2. 2

   Measured gradient 0.98 is within ~2 % of ideal.

   $\text{percentage deviation} = 2\%$

3. 3

   Comment on uncertainty.

   $\pm 0.5^\circ \text{ at } 30^\circ \Rightarrow \pm 1.7\%$

Solution

1. 1

   Recall the wave equation.

   $v = f \lambda$

2. 2

   Rearrange for wavelength.

   $\lambda = \dfrac{v}{f}$

3. 3

   Substitute the values for water.

   $\lambda = \dfrac{1500}{256} \approx 5.86 \text{ m}$

4. 4

   Comment on why this differs from the wavelength in air. Sound travels faster in water, so for the same frequency the wavelength must be larger.

   $\lambda_{\text{water}} > \lambda_{\text{air}} \text{ because } v_{\text{water}} > v_{\text{air}}$

Solution

1. 1

   Recall the human hearing range.

   $\text{20 Hz to 20 000 Hz}$

2. 2

   Compare the whistle's frequency with the upper limit.

   $30\,000 \text{ Hz} > 20\,000 \text{ Hz}$

3. 3

   Conclude that the whistle is above the human range (ultrasound) but inside a dog's range (dogs hear up to ~45 kHz).

Solution

1. 1

   State the nature of sound waves.

   $\text{sound is longitudinal: compressions and rarefactions}$

2. 2

   State what sound requires.

   $\text{sound requires a medium of particles}$

3. 3

   Apply the conditions on the Moon.

   $\text{Moon has effectively no atmosphere} \Rightarrow \text{no medium}$

4. 4

   Conclude. The meteor's impact may produce sound in the ground, but no air carries it to her ear, so nothing reaches the eardrum.

Solution

1. 1

   Identify the round-trip time and the wave speed.

   $t = 40\,\mu\text{s} = 40\times10^{-6}\,\text{s},\; v = 1540\,\text{m/s}$

2. 2

   Use $s = vt$ to find the total path length.

   $s_{\text{total}} = 1540 \times 40\times10^{-6} = 0.0616\,\text{m}$

3. 3

   Halve to get depth (the pulse travelled there and back).

   $\text{depth} = \dfrac{0.0616}{2} = 0.0308\,\text{m}$

4. 4

   Convert to a sensible unit.

   $0.0308\,\text{m} = 3.08\,\text{cm} \approx 31\,\text{mm}$

Solution

1. 1

   Recall the echo formula.

   $\text{depth} = \dfrac{v\,t}{2}$

2. 2

   Substitute.

   $\text{depth} = \dfrac{1500 \times 4.0}{2}$

3. 3

   Evaluate.

   $\text{depth} = 3000\,\text{m} = 3\,\text{km}$

Solution

1. 1

   State the property of S-waves that matters.

   $\text{S-waves are transverse and cannot travel through liquids}$

2. 2

   State what is observed on the far side of the Earth.

   $\text{P-waves arrive but S-waves do not}$

3. 3

   Conclude using the simplest model that fits.

   $\text{at least one inner layer must be a liquid that absorbs S-waves}$

4. 4

   Identify the layer (outer core).