A longitudinal wave of amplitude 3.0 cm, frequency 2.0 Hz , and speed 3.8 m/s travels on an infinitely long Slinky.

How far apart (the closest distance) are the two nearpoints on the Slinky that at one particular time both have the maximum displacements from their equilibrium positions?

Answer:

λ= 1.9 m

Explanation:

Given Data

Amplitude=3.0 cm

frequency=2.0 Hz

Speed=3.8 m/s

To find

Distance=?

Solution

As the wavelength is a measure of distance between two Crest or Troughs in a wave the peaks should be identical.

So

λ(wavelength) = v(speed) / f(frequency)

λ= 3.8m/s / 2.0/s

λ= 1.9 m

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-12-09T11:02:00+01:00

The closest distance between the two nearpoints on the Slinky is 1.9 m.

The given parameters:

amplitude of the wave, A = 3.0 cm
frequency of the wave, f = 2.0 Hz
speed of the wave, v = 3.8 m/s

The closest distance between the two nearpoints on the Slinky is calculated by applying wave equation as follows;

v = fλ

[tex]\lambda = \frac{v}{f} \\\\\lambda = \frac{3.8}{2} \\\\\lambda = 1.9 \ m[/tex]

Thus, the closest distance between the two nearpoints on the Slinky is 1.9 m.

The complete question is below;

A longitudinal wave of amplitude 3.0 cm, frequency 2.0 Hz , and speed 3.8 m/s travels on an infinitely long Slinky. How far apart (the closest distance) are the two nearpoints on the Slinky that at one particular time both have the maximum displacements from their equilibrium positions?

Learn more about wave equation here: https://brainly.com/question/10478236