Sound Worked ExamplesIGCSE Physics 0625

Because the time-base is the same, more complete waves across the screen means a higher frequency. Count the waves: A $3$, B $5$, C $3$, D $2$.

(a) Pitch depends on frequency, so the highest pitch is B.

(b) Loudness depends on amplitude — the height of a peak above the centre line. C has the largest amplitude ($2.5$ divisions), so C is the loudest.

(c) C has the same number of waves as A (same frequency, so same pitch) but a larger amplitude: C.

The approximate range of human hearing is $20$ Hz to $20\,000$ Hz ($20$ kHz).

Both can hear a frequency only if it is inside both ranges. The overlap starts at the higher of the two lower limits ($20$ Hz) and ends at the lower of the two upper limits ($12$ kHz): B.

A combines the two ranges instead of overlapping them. C is the elephant's whole range, but humans cannot hear $16$ Hz to $20$ Hz. D is the band that humans can hear and elephants cannot.

(a) Sound is a longitudinal wave: each particle vibrates backwards and forwards parallel to the direction in which the sound (the energy) travels. The particles do not travel along with the wave.

(b) At J the particles are close together: J is a compression (higher pressure). At K they are far apart: K is a rarefaction (lower pressure).

(c) One wavelength is the distance between the centres of two neighbouring compressions. The compressions are at $0.4$ m and $1.2$ m.

(d) Use the wave equation.

This lies in the expected range for air, $330$–$350$ m/s.

(a) Ultrasound is sound with a frequency higher than $20$ kHz (above the range of human hearing).

(b) Start by writing the equation in symbols, then rearrange it.

Convert the frequency to hertz and substitute.

Evaluate to one more significant figure than the value you are asked to show.

In general, sound travels faster in liquids than in gases, so the speed increases in water. This rules out C and D.

The frequency is set by the vibrating source, so it stays the same when the wave crosses into a new medium. This rules out A.

From $v = f\lambda$ with $f$ fixed, a larger $v$ needs a larger $\lambda$: the wavelength increases. The answer is B.

C is the pattern for light entering glass — light slows down in a denser material, but sound speeds up.

(a) Convert the distance to metres.

Rearrange speed $=$ distance ÷ time.

Round correctly to 2 significant figures.

(b) Light travels at $3.0 \times 10^{8}$ m/s, so it takes only $1900 / (3.0 \times 10^{8}) = 6.3 \times 10^{-6}$ s — about a millionth of the sound's travel time, far too small to notice.

(a) Marking points: stand a large distance (at least $50$ m) from a large, flat wall and measure the distance $d$ with a trundle wheel or long measuring tape.

Make a short, loud sound — clap. Adjust the clapping rate until each clap happens at the same moment as the echo of the previous one.

The timer uses a stopwatch to time a large number of intervals (for example $20$), then divides by the number of intervals. Timing many intervals reduces the effect of reaction time.

In one interval the sound travels to the wall and back, a distance of $2d$. Speed $= 2d \div$ (time for one interval). Repeat and average.

(b) Time for one interval, then the total path length.

Calculate the speed.

(a) Marking points: a pulse of ultrasound is sent down into the water; it reflects from the seabed (an echo); the time between sending the pulse and receiving the echo is measured; distance travelled $=$ speed $\times$ time, and the depth is half of this because the pulse goes down and back.

(b) Total distance to the fish and back, then halve it.

(c) The same method with the seabed echo time.

The echo is the sound reflected from the cliff, so the sound travels to the cliff and back: a total distance of $2d$.

Time $=$ distance ÷ speed, so $t = 2d / v$: B.

A is the time to reach the cliff only. C is upside down (its unit is 1/s, not s). D has the unit m²/s — checking units quickly rules out C and D.

(a) The sound reaches her along two paths: through the steel and through the air. Sound travels much faster in solids than in gases, so the sound through the steel arrives first and the sound through the air arrives later.

(b) Time through the steel.

Time through the air.

Subtract.

(a) Ultrasound is a longitudinal sound wave of frequency above $20$ kHz. It travels through the solid steel and reflects from any boundary inside, such as a crack, so a crack shows up as an extra echo without damaging the block (non-destructive testing).

(b) Measure from the transmitted pulse to the crack echo: $3.0$ divisions.

The pulse travels down to the crack and back, so halve the distance.

(c) The far-side echo is $8.0$ divisions after the transmitted pulse.

(a) Use a large triangle on the line. It crosses the time axis at $t = 0.10$ s ($d = 0$) and passes through $t = 1.60$ s, $d = 510$ m.

The gradient is distance ÷ time, so it is the speed of sound in air: $340$ m/s.

(b) Every time reading is about $0.10$ s too long. A likely reason: the student is slower to react to the bang (stopping the watch) than to the flash (starting it), adding a roughly constant delay to every reading — a systematic error.

(c) A constant extra time shifts every point sideways by the same amount. The whole line moves along the time axis but its slope is unchanged, so the gradient — the speed — is still correct. Calculating $d/t$ for a single reading would not remove the delay (for example $130 / 0.48 = 271$ m/s).