. a string has a mass of 12.5 g. the string is stretched with a force of 9.63 n, giving it a length of 1.89 m. then, the string vibrates transversely at precisely the frequency that corresponds to its 4th normal mode. what is the frequency of the standing wave? (a.) 42.9 hz (b.) 25.9 hz (c.) 32.6 hz (d.) 40.4 hz (e.) none of the above



Answer :

The frequency of the standing wave is  1.16 hertz.So option (e) is correct.

What is standing wave?

standing wave, also called stationary wave, combination of two waves moving in opposite directions, each having the same amplitude and frequency. The phenomenon is the result of interference; that is, when waves are superimposed, their energies are either added together or canceled out. In the case of waves moving in the same direction, interference produces a traveling wave. For oppositely moving waves, interference produces an oscillating wave fixed in space.

A vibrating rope tied at one end will produce a standing wave, as shown in the figure; the wave train (line B), after arriving at the fixed end of the rope, will be reflected back and superimposed on itself as another train of waves (line C) in the same plane. Because of interference between the two waves, the resultant amplitude (R) of the two waves will be the sum of their individual amplitudes. Part I of the figure shows the wave trains B and C coinciding so that standing wave R has twice their amplitude.

According to question:

m = 12.5 g, L = 1.89 m, F = 9.63N , n = 4

Using formula;

v = √F/μ , μ = m/L = 12.5/1.89 = 6.613

v = √9.63/6.61 = √1.5

Then,

Frequency(f) = nv/2L

f = 4(√1.5)/2(1.89)

f = 4.4/3.78 = 1.16 hertz

Thus required frequency is 1.16 hertz.

To learn more about amplitude refer to:

brainly.com/question/3613222

#SPJ4