435-Hz sound or a 441-Hz sound. It is producing a 441-Hz sound and could be prod

Question

435-Hz sound or a 441-Hz sound. It is producing a 441-Hz sound and could be producing no other sound frequency. Transverse Standing Waves Two transverse standing waves are shown in the drawing. The strings have the same tension and length, but the bottom string is more massive. Which standing wave, if either, is vibrating at the higher frequency? :The top standing wave has the higher frequency, because the traveling waves have a smaller speed due to the smaller mass of the string. The top standing wave has the higher frequency, because the traveling waves have a larger speed due to the smaller mass of the string Both standing waves have the same frequency, because the frequency of vibration does not depend on the mass of the string. The bottom standing wave has the higher frequency, because the traveling waves have a smaller speed due to the larger mass of the string The bottom standing wave has the higher frequency, because the traveling waves have a larger speed due to the larger mass of the string. A standing wave on a string fixed at both ends is vibrating at its fourth harmonic. If the length, tension, and finer density are kept constant, what can be said about the wavelength and frequency of the fifth harmonic relative to the fourth harmonic? The wavelength

Explanation / Answer

Frequency f = v / 2L where   v = speed of the wave             L = length of the string from above equation f ' / f = v ' / v                                         = [ T ' / m ' ] / [ T / m]     since speed v = [ T / m]                                         = [ (T ' / m ') *( m / T ) ]                                         = [ m / m ' ]     Since tension is same i.e., T ' = T where m = linear density of the string  = M / L So, f ' / f = [ M / M ' ]       Since L is same bottom is more massive.So, M ' > M Therefore f ' / f < 1                       f ' < f i.e., frequency of lower string is smaller than upper string                           Therefore option ( b) is correct

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