Equations to use: v= λ ∙ f v=d/t

b. A guitar string has a length of 0.90 m. When you pluck it, it plays a “C” that has a frequency of 256 Hz. How fast is the wave moving back and forth along the string? (

c. In a train track made of solid steel, sound waves travel extremely fast – about 6,000 m/s. If a train’s vibration travels for 3 seconds, how far will it move?

We have a stringth with length L = 0.90 m, and a standing wave with frequency f = 256 Hz. The equation that relates the speed of the wave with the frequency and the length of the string is

[tex]f=\frac{v}{2L}[/tex]

where v is the speed of the wave. Re-arranging the equation and putting in the numbers, we find:

[tex]v=2Lf=2(0.90 m)(256 Hz)=460.8 m/s[/tex]

c. 18,000 m

Explanation:

The speed of the sound wave in steel is v = 6000 m/s. The train's vibration travels for 3 seconds, so the distance covered by the sound wave during this time can be calculated using the formula

[tex]v=\frac{d}{t}[/tex]

Re-arranging the equation and substituting the numbers, we can find d:

[tex]d=vt=(6000 m/s)(3 s)=18,000 m[/tex]

zlskfrum zlskfrum · Answer 2 · 2019-03-06T04:17:53+01:00

zlskfrum zlskfrum

06-03-2019

b. The guitar represents half of the wave length. So the full wave length is 2x0.9m = 1.8m.

Using the given equation, v= λ ∙ f,

the wave is moving back and forth along the string at 1.8 ∙ 256

= 460.8m/s

c. Sound waves travel at 6,000 m/s.

Using the given equation, v=d/t, or d=vt,

train's vibration in 3 seconds travels 6000*3

=18,000m or 18km

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