Waves, Sound & the Electromagnetic Spectrum

8 methods · 25 questions

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A pulse that reflects and returns, or two pulses from one event, lets you read a distance straight from the timing, provided you insert the right factor: a reflected pulse covers twice the one-way distance, and two signals of different speed over the same path separate by a time gap set by the difference of their travel times.

Trigger: an echo, a reflected ultrasound or radar pulse, a crack-and-far-end scan, or two waves of different speed from one source.

Instances:
(i) reflection depth is $d = v t /2$ , halve the measured there-and-back time;
(ii) a source offset by $x$ from a midpoint adds $2 x$ to one path, so $2 x = v Δ t$ ;
(iii) the earlier returning pulse is always the nearer reflector, so subtract the two depths to get a gap;
(iv) two speeds over one distance give $d (\frac{1}{v _{slow}} - \frac{1}{v _{fast}}) = Δ t$ ;
(v) an echo coincides with the emission when its delay $2 d / v$ is a whole-number multiple of the pulse period.

Linked questions (3)

NSAA 2016 Section 1 Part B (ESAT Physics) · Q36

A student carries out an experiment to measure the speed of sound. A loudspeaker that emits sound in all directions is placed between two buildings that are

128 m

apart. The student and loudspeaker are

48 m

from one of the buildings.

The loudspeaker is connected to a signal generator that causes it to emit regular clicks. The student notices that each click results in two echoes, one from each building. The rate at which the clicks are produced is gradually increased from zero until each echo coincides with a new click being emitted by the loudspeaker.

What is the frequency of emission of clicks when this happens?

(The speed of sound in air

= 320 m s^{- 1}

A diagram of two buildings facing each other, 128 m apart, drawn as thick horizontal lines seen edge-on with a small vertical wall at each end. A dot labelled 'student and loudspeaker' sits on the line between them, 48 m from the left-hand building (and therefore 80 m from the right-hand building). — illustrative

A $2.0 Hz$
B $2.5 Hz$
C $3.3 Hz$
D $4.0 Hz$
E $5.3 Hz$
F $6.7 Hz$
G $10 Hz$

Worked solution

Answer:G

The student and loudspeaker are

48 m

from one building and

128 - 48 = 80 m

from the other.

The echo from the near building returns after travelling there and back: time

= \frac{2 \times 48}{320} = 0.30 s

. The echo from the far building returns after time

= \frac{2 \times 80}{320} = 0.50 s

.

For an echo to coincide with a new click, the echo's delay must be a whole number of click periods

T

: so

0.30 s

and

0.50 s

must both be whole-number multiples of

T

. As the click rate is increased from zero (so

T

decreases from a large value), the first time this happens simultaneously for both echoes is at the largest

T

that divides evenly into both

0.30

and

0.50

: this is

T = 0.10 s

(since

0.30 = 3 \times 0.10

and

0.50 = 5 \times 0.10

, and no larger common value works, as

0.30

's divisors

0.30, 0.15, 0.10, \dots

and

0.50

's divisors

0.50, 0.25, 0.1 \overline{6}, 0.125, 0.10, \dots

first coincide at

0.10

).

The frequency is

f = \frac{1}{T} = \frac{1}{0.10} = 10 Hz

.

The answer is G.

NSAA 2017 Section 1 Part B (ESAT Physics) · Q24

In one type of medical scanner a source is placed inside a patient's body. This source causes pairs of gamma-rays to be emitted simultaneously in opposite directions.

Detectors on each side of the patient are used to detect the gamma-rays. The distance between the two detectors is

3.0 m

. When the source is at Q, half-way between the detectors, the two gamma-rays arrive at the same time.

In a particular scan the gamma-rays arrive at the two detectors with a time difference of

4.0 \times 1 0^{- 10} s

.

Assume that, inside the patient, the gamma-rays travel at a speed of

3.0 \times 1 0^{8} m s^{- 1}

.

How far from Q, half-way between the detectors, is the gamma-ray source?

A $6.0 mm$
B $12 mm$
C $24 mm$
D $6.0 cm$
E $12 cm$
F $24 cm$

Worked solution

Answer:D

If the source sits a distance

x

from the midpoint Q (toward one detector), it is

x

closer to that detector and

x

farther from the other, compared with being at Q. The extra path length the far gamma-ray must travel, compared with the near one, is therefore

2 x

, and this extra distance is what produces the arrival time difference:

2 x = v Δ t = 3.0 \times 1 0^{8} \times 4.0 \times 1 0^{- 10} = 0.12 m .

x = 0.060 m = 6.0 cm

.

The answer is D.

NSAA 2021 Section 1 Part B (ESAT Physics) · Q40

A pulse of ultrasound travels from one end of a solid uniform rod of length

L

, starting at time

t = 0

.

The pulse is partially reflected by a crack in the rod and partially by the far end of the rod.

These two reflected pulses travel back along the rod, arriving at the end from which they started at times

t_{1}

and

t_{2}

, where

t_{2} > t_{1}

.

What is the distance between the crack and the far end of the rod?

A $\frac{t _{1}}{t _{2}} L$
B $\frac{t _{2}}{t _{1}} L$
C $\frac{t _{1}}{2 t _{2}} L$
D $\frac{t _{2}}{2 t _{1}} L$
E $\frac{( t _{2} - t _{1} )}{t _{2}} L$
F $\frac{( t _{2} - t _{1} )}{2 t _{2}} L$

Worked solution

Answer:E

Let

v

be the (constant) speed of ultrasound in the rod, and

d

the distance from the starting end to the crack.

The pulse reflected by the crack travels a total distance

2 d

(there and back) in time

t_{1}

2 d = v t_{1} ⟹ d = \frac{v t _{1}}{2} .

The pulse reflected by the far end travels a total distance

2 L

in time

t_{2}

2 L = v t_{2} ⟹ v = \frac{2 L}{t _{2}} .

The distance between the crack and the far end is

L - d = L - \frac{v t _{1}}{2} = L - \frac{1}{2} (\frac{2 L}{t _{2}}) t_{1} = L - \frac{L t _{1}}{t _{2}} = L (1 - \frac{t _{1}}{t _{2}}) = \frac{( t _{2} - t _{1} )}{t _{2}} L .

The answer is E.