# Chapter 13 - Worked Solutions

### to questions from the OUP text *Senior Physics -
Concepts in Context* by Walding,
Rapkins and Rossiter

My thanks to Peter Finch from St Joseph's College, Gregory Terrace, for these (pdf) solutions.

Solutions-Ch-13-Finch.pdf

The rest are by Greg Rapkins:

**Q37.**

The sound that is reflected from the stand has to travel 300 m further then the other sound.

Therefore

v = d/t = 300m /0.9 s = 333 m s^{-1}

**Q38.**

In general: L =
nl/4, where n = 1, 2, 3,
4, 5, .......

**Q39.**

Sound is reflected from the fish and from the bottom.

Since the sound has traveled to the fish (and bottom) and back. It took 0.1 s to reach the fish and 0.125 s to reach the bottom. That is the sound took 0.025 s to travel from the fish to the bottom.

v = d/t

d = v.t

= 1450 x .025 = 36.25 m

the fish are 36 m from the bottom.

**Q41.**

Solutions-Ch13-Q41-Finch.pdf

**(a)**

Points A, B, C, are points on the spring and therefore are moving outwards or inwards.

When the wave has gone 1 m further on point A is the same distance form the equilibrium.

Point B will have moved up to about the position o the 18 cm mark - a change of 38 cm.

Point c has moved soen to the -40 cm mark - a change of about -20 cm .

Therefore B has moved the greatest distance in the same time and therefore has the greatest velocity.

**(b)**

The velocity of the wave is 5 m s^{-1} and therefore goes 1 m in 0.2 of a second. When the wave has gone 1 m, C has moved down 20 cm.

Speed = distance / time = 20 cm/0.2 s = 100 cm s^{-1} = 1 m s^{-1}.

**(c)**

**(d)**

Points have to instantaneously change direction.

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