Question

A distance of 2.60 meters lies between two successive minima of a transverse wave. Four wave crests pass a stationary observer every 11.7 seconds. (a) Compute the wave's frequency (in Hz). (No Response) Hz (b) With what speed (in m/s) does the wave travel? (No Response) m/s

Answer 1

Solution) Lambda = 2.60 m

4 crests pass for every 11.7 s

(a) Frequency, f = ?

f = (4)/(11.7) = 0.3418

Approximately , f = 0.342 Hz

(b) speed , V = ?

V = f(Lambda)

V = 0.342×2.60

V = 0.889 m/s