Set 2: Wave Speed
Problems pg 443 #56,57,59,61,62,63,64

56.

• 1.5 m/s

Use v_w = fl where l = 2.4 meters and the period, T = 1.6 seconds. Remember that f = 1/T.

v_w = (1/1.6)(2.4)
v_w = 1.5 m/sec

57.

• 3 x 10^-6 m

Use v_w = fl where v_w = 300,000 km/sec = 3 x 10⁸ m/sec and the frequency = 10¹⁴ hz.

v_w = fl
3 x 10⁸ = 10¹⁴ l
l = 3 x 10⁸ / 10¹⁴
l = 3 x 10^-6 m

59.

• 1.73 cm to 17.3 m

Use v_w = fl where v_w = 345 m/sec and f = 20 hz and then f = 20,000 hz.

v_w = fl
345 = 20l
l = 345 / 20 = 17.3 meters

v_w = fl
345 = 20000 l
l = 345 / 20000 = 0.0173 meters = 1.73 cm

61.

• 1.00 x 10⁴ m

Use v_w = fl where l = 4.80 x 10^-4 meters and f = 2.50 x 10⁶ hz

v_w = fl
v_w = (2.50 x 10⁶ )(4.80 x 10^-4 )
v_w = 1200 m/sec

Since the echo must travel both down to the bottom and then back up to the ship, it only takes 16.7 / 2 or 8.35 seconds for a one-way trip.

Using d = rt
d = (1200)(8.35)
d = 10020 meters or 1.00 x 10⁴ meters

62.

• 3.0 m/sec

13 crests translates into 12 wavelengths, therefore the frequency = 12 waves/3 seconds = 4 hz

Use v_w = fl where l = 0.75 meters and f = 4 hz
v_w = (4)(0.75)
v_w = 3.0 m/sec

63.

• 0.15 m
• 15 cm/s

The amplitude should be read from the second graph. Note that the height of a crest or the depth of a trough is 15 cm = 0.15 meters. The wavelength should be read from the first graph. Since the first crest is at 3.0 cm and the second crest is at 15.0 cm, the wavelength equals 15 - 3 = 12.0 cm. The period should be read from the second graph. Since the first crest is at 0.20 sec and the second crest is at 1.0 sec, the period equals 1.0 - 0.2 = 0.8 seconds. Use vw = fl where l = 0.12 meters and f = 1/0.8 = 1.25 hz vw = (1.25)(0.12) vw = 0.15 m/sec

64.

• 0.20 m
• 0.13 m increase

Use v_w = fl where v_w = 12 m/sec and f = 60 hz
12 = 60l₁
l₁ = 12/60
l₁ = 0.20 meters

Use v_w = fl where v_w = 20 m/sec and f = 60 hz
 20 = 60l₂
l₂ = 20/60
l₂= 0.33 meters

Dl = l₂ - l₁ = 0.33 - 0.20 = 0.13 meters