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?

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

b. 460.8 m/s

Explanation:

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

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

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

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

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft