7. Waves

A 5.0 Hz source has period 0.20 s. Two points separated by λ/2 are 180° (π rad) out of phase; points separated by one wavelength are in phase.

One cycle spans 4.0 divisions at 2.0 ms div⁻¹: T = 8.0 ms and f = 125 Hz. A peak 3.0 divisions above the centre at 0.50 V div⁻¹ has amplitude 1.5 V.

v is propagation speed in m s⁻¹, f is source frequency in Hz and λ is wavelength in m. The derivation tracks the travelling pattern, not the local speed of an oscillating particle.

If the medium fixes v, a change in source frequency changes wavelength inversely. Use metres and hertz so the result is m s⁻¹.

If a wave travels at 12 m s⁻¹ with frequency 4.0 Hz, its wavelength is 3.0 m.

In a mechanical wave, neighbouring parts of the medium interact: local oscillations pass energy onward while particles have no overall journey with the wave. A water ripple can make a floating cork oscillate and deliver energy to the edge.

A material medium is not required for every progressive wave. Electromagnetic waves transfer energy through vacuum as well as through materials.

A 6.0 W wave spread uniformly over 3.0 m² has intensity 2.0 W m⁻². Under otherwise unchanged conditions, doubling amplitude makes intensity four times larger; halving amplitude makes it one quarter.

Both types can be progressive waves that transfer energy, obey v = fλ, reflect, refract, diffract, interfere and form stationary waves. These shared behaviours do not determine whether a wave is transverse or longitudinal.

For a wave travelling to the right, the instantaneous particle velocity has the opposite sign to the local slope of a displacement–distance graph: positive slope means negative particle velocity, and negative slope means positive particle velocity. At maximum or minimum displacement, instantaneous particle velocity is zero.

Interpret in this order: read both axis labels and units; decide whether the graph is a time record or a spatial snapshot; identify the stated positive displacement and propagation directions; then use spacing, gradient and phase. Equivalent-phase spacing gives λ only on a distance axis and T only on a time axis.

The moving source emits successive wavefronts from different positions. In the same medium, sound speed remains v, so the changed wavelength gives a changed arrival frequency through v = fλ. The source itself can continue emitting at one steady frequency.

Greater speed towards the observer gives a greater upward shift; greater speed away gives a greater downward shift. A siren therefore sounds higher before passing and lower after passing a stationary listener.

A horn emits 440 Hz while moving towards a stationary observer at 30.0 m s⁻¹; sound speed is 340 m s⁻¹. f_o = 440 × 340/(340 − 30.0) = 483 Hz (3 s.f.). The answer is above 440 Hz, as approach requires.

First predict higher or lower frequency. Then select the sign, substitute speeds in the same units, solve, and compare f_o with f_s. The same equation can be rearranged to find v_s; retain the physical sign and direction.

They do not require a material medium; in matter their speed can be lower and depends on the medium.

Radio, visible light and X-rays are different frequencies of the same transverse electromagnetic family.

Convert first, then classify. Useful anchors are 1 mm = 10⁻³ m, 1 μm = 10⁻⁶ m and 1 nm = 10⁻⁹ m. Thus 2.1 cm = 2.1 × 10⁻² m is microwave, 12 μm = 1.2 × 10⁻⁵ m is infrared, and 138 pm = 1.38 × 10⁻¹⁰ m is X-ray.

If frequency is given, calculate λ = c/f before using the table. For f = 3.0 × 10¹⁶ Hz, λ = 1.0 × 10⁻⁸ m, at the approximate ultraviolet/X-ray boundary; use the convention and options supplied by the question.

Treat the limits as approximate and use wavelength/frequency trends when identifying colour or neighbouring ultraviolet and infrared.

A 500 nm wave lies in the visible range, while 350 nm is ultraviolet and 900 nm is infrared.

A polarising filter has a transmission axis and transmits the component of the incident electric-field oscillation along that axis. A longitudinal wave oscillates parallel to propagation, so it has no set of transverse directions from which one can be selected; it cannot be polarised.

Rotating a filter in plane-polarised light changes the transmitted intensity and can reduce it to zero when the transmission axis is perpendicular to the incident polarisation direction. Observing polarisation is therefore evidence that a wave is transverse.

For each filter in order: find θ from the incoming polarisation direction to the filter axis; calculate I = I₀ cos²θ; use this transmitted I as the next filter's I₀; and set the new polarisation direction equal to the axis of the filter just passed.

Vertically plane-polarised light of intensity 8.0 W m⁻² meets a first axis at 50° and a second axis at 20° to vertical. First, I₁ = 8.0 cos²50° = 3.31 W m⁻². The axes differ by 30°, so I₂ = 3.31 cos²30° = 2.48 W m⁻² ≈ 2.5 W m⁻².