Question 4: Mac and Tosh stand 8 meters apart and demonstrate the motion of a transverse wave on a string. The wave can be described as having a vertical
distance of 32 cm from a trough to a crest, a frequency of 2.4 Hz, and a
horizontal distance of 48 cm from a crest to the nearest trough. Determine the
amplitude, wavelength, and speed of such a wave. Show all of your work and
explain your steps for solving the problem

Explanation:

Refer to the pic below that I attached as we talk through these answers.

The amplitude is shown as the height from the top of one wave (called the crest) to the midline. If the whole vertical height from crest to trough is 32 cm, then the amplitude is half of that at

A = 16 cm

The wavelength, or λ, is the horizontal distance from crest to crest or trough to trough. Therefore, our wavelength is 48 cm + 48 cm which is

λ = 96 cm

The speed of the wave is the same thing as the velocity of the wave which can be found in the equation

where f is the frequency of the wave and lambda is the wavelength. Filling in:

and

2.4(96) = v so

v = 230 m/s to the correct number of sig fig's

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form