Sown below are 5 successive snapshots of a standing wave on a vibrating string t

Question

Sown below  are 5 successive snapshots of a standing wave on a vibrating string that is 0.6[m] long.  The time interval between snapshots is (1/800)[s].  The speed of the traveling waves on the string is v=lambda(f)=?.  (the amplitude of the wave above and below the resting position is 1.5cm.

Sown below are 5 successive snapshots of a standing wave on a vibrating string that is 0.6[m] long. The time interval between snapshots is (1/800)[s]. The speed of the traveling waves on the string is v=lambda(f)=?. (the amplitude of the wave above and below the resting position is 1.5cm.

Explanation / Answer

time period of the wave= 5*1/800= 1/160sec

freq.=1/T=160 Hz

reqd. vel of wave= lambda*f=0.6*160=96 m/sec

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