Calculate the distance travelled by a sound wave having frequency 1000 Hz and wavelength 0·25 m, if it travels for 5 seconds in a certain medium.

Question

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Correct — option (2), 1250 m. The problem is solved in two short steps and the only thing that can go wrong is the handling of units. The first step uses the wave equation, which states that the speed of a wave equals its frequency multiplied by its wavelength. Here the frequency is 1000 hertz, meaning a thousand complete oscillations pass a point every second, and the wavelength is 0·25 metre, printed in this paper with a raised middle dot in place of the usual decimal point. Multiplying gives a speed of 1000 times 0·25, which is 250 metres per second. The physical picture behind the multiplication is worth holding, because it makes the formula impossible to invert by mistake: in one second the source completes a thousand cycles and therefore emits a thousand wavelengths, and since each wavelength occupies a quarter of a metre, the leading edge of the disturbance has advanced a thousand quarters of a metre, which is 250 metres. Frequency times wavelength is a distance per second, and a distance per second is a speed. The second step is ordinary uniform motion. The wave travels for 5 seconds at 250 metres per second, so it covers 250 multiplied by 5, which is 1250 metres, and that is option (2). Two checks confirm the answer without repeating the calculation. The first is dimensional: hertz is per second and metre is a length, so their product is a length per second, and multiplying by a time in seconds returns a length in metres, which is what the question asks for. The second is physical, and it disposes of two of the wrong options at once: a speed of 250 metres per second is a reasonable value for sound in a gaseous medium, being of the same order as the familiar figure of about 340 metres per second for air, whereas an answer measured in tens of kilometres over five seconds would require speeds of ten or sixteen kilometres per second, far above the speed of sound in any ordinary material — sound travels at roughly fifteen hundred metres per second in water and at a few thousand in steel. Note that the stem specifies a certain medium without naming it, which is precisely why the frequency and the wavelength are supplied: the speed is not to be recalled but to be computed from them. A candidate who substitutes a remembered speed of sound in air for the value the data gives will arrive at a different number and will not find it among the options. Option (2) is the answer.

Why the other options are wrong — (a) 50 km is 50000 metres, and covering that distance in 5 seconds would require a speed of 10000 metres per second, or ten kilometres per second. Nothing in the data yields such a figure: the frequency and the wavelength give 250 metres per second, and no rearrangement of 1000 and 0·25 produces ten thousand. The option is one of two in this set expressed in kilometres rather than metres, and its function is to catch a candidate who has mishandled a unit conversion somewhere in the working and is then reassured to find a kilometre answer waiting for him. It can also be rejected on physical grounds alone, before any arithmetic: ten kilometres per second is far faster than sound travels in any ordinary medium, being several times the speed of sound in steel and many times its speed in water or air, so a wave described simply as a sound wave in a certain medium cannot be moving at that rate.  (c) 800 m would mean a speed of 160 metres per second over the 5 seconds, and that does not follow from a frequency of 1000 hertz and a wavelength of 0·25 metre, which give 250 metres per second. The option is the most dangerous of the three because it is expressed in metres, is of the same order of magnitude as the correct answer, and corresponds to a speed that is at least physically conceivable for sound in a gas, so neither the unit check nor the plausibility check will reject it. Only the arithmetic will. This is the standard reason for computing rather than estimating in a numerical question: three of the four options can often be dismissed by inspection, and the fourth is always placed close enough to the answer that inspection cannot separate them.  (d) 80 km is 80000 metres, which over 5 seconds implies a speed of 16000 metres per second, or sixteen kilometres per second. Like the other kilometre option it fails both the arithmetic and the physics: the data give 250 metres per second, and sixteen kilometres per second exceeds the speed of sound in any common material by a wide margin, since even steel carries sound at only a few thousand metres per second. The presence of two kilometre options among four is itself informative about how the set was built, because it means that a candidate who has made a unit error of any kind has a fifty per cent chance of finding a home for it. The discipline that prevents this is to carry the units through every line of the working — hertz times metre gives metres per second, and metres per second times seconds gives metres — so that the answer arrives already labelled.

Concept: The wave equation is the single relation that connects the three quantities used to describe any periodic wave, and it says that the speed equals the frequency multiplied by the wavelength. Each of the three has a distinct physical origin and understanding which is fixed by what makes the relation easy to use rather than easy to misremember. The frequency is determined by the source: a tuning fork, a vocal cord or a loudspeaker vibrating a thousand times a second sends out a thousand cycles a second, and that number does not change as the wave moves from one medium into another. The speed is determined by the medium, through its elasticity and its density, which is why sound travels at roughly three hundred and forty metres per second in air at ordinary temperature, about fifteen hundred metres per second in water, and several thousand in a solid such as steel. The wavelength is then whatever the other two require it to be, and it is the quantity that changes when a wave crosses from one medium into another while its frequency stays the same. Sound is a longitudinal mechanical wave, propagating as a train of compressions and rarefactions and requiring a material medium; it cannot travel through a vacuum. The audible range for human beings runs from about twenty hertz to about twenty thousand hertz, so the thousand hertz in this problem sits comfortably inside it. Problems built on the wave equation come in three forms — give two quantities and ask for the third; combine the equation with uniform motion to obtain a distance or a time, which is what this question does; or combine it with reflection to obtain the distance of a reflecting surface from the time of an echo, remembering in that case to halve the total path. In all three, the working is short and the risk lies entirely in the units, so the habit of writing the unit beside every number and cancelling as one goes is worth more than any amount of additional formula learning.

State PCS 2022 Q40 — Science & Technology PYQ with Answer

State PCS 2022 Q40 — Science & Technology PYQ with Answer