Question

1. a hiker shouts toward a cliff that is a distance of 685 meters away. the shout is heard on the other side 4.00 s of time later.

a) what is the speed of sound?
b) if the wavelength is.750 m, what is the frequency?
c) what is the period of this vibration?

Explanation:

Step1: Calculate the total distance traveled by sound

The sound travels to the cliff and back. So the total distance \(d = 2\times685\space m=1370\space m\)

Step2: Find the speed of sound

Using the formula \(v=\frac{d}{t}\), where \(t = 4.00\space s\) and \(d = 1370\space m\). Then \(v=\frac{1370}{4}=342.5\space m/s\)

Step3: Calculate the frequency

Using the formula \(v = f\lambda\), where \(v = 342.5\space m/s\) and \(\lambda=0.750\space m\). Then \(f=\frac{v}{\lambda}=\frac{342.5}{0.750}\approx456.7\space Hz\)

Step4: Calculate the period

Using the formula \(T=\frac{1}{f}\), where \(f\approx456.7\space Hz\). Then \(T=\frac{1}{456.7}\approx0.0022\space s\)

Answer:

a) The speed of sound is \(342.5\space m/s\)
b) The frequency is approximately \(456.7\space Hz\)
c) The period is approximately \(0.0022\space s\)