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 speed of sound

The sound travels to the cliff and back. So total distance \(d = 2\times685\space m=1370\space m\), time \(t = 4.00\space s\). Using formula \(v=\frac{d}{t}\), we have \(v=\frac{1370}{4}\).

Step2: Calculate frequency

Using formula \(v = f\lambda\), where \(v\) is speed of sound (from part a), \(\lambda=0.750\space m\). Then \(f=\frac{v}{\lambda}\).

Step3: Calculate period

Using formula \(T=\frac{1}{f}\), where \(f\) is frequency (from part b).

Answer:

a) \(v=\frac{2\times685}{4}=342.5\space m/s\)
b) \(f=\frac{342.5}{0.750}\approx456.7\space Hz\)
c) \(T=\frac{1}{456.7}\approx0.0022\space s\)