In the arrangement shown in the figure below. an object of mass 7.1 kg hangs fro

Question

In the arrangement shown in the figure below. an object of mass 7.1 kg hangs from a cord around a light pulley. The length of the cord between point P and the pulley is 2.7 m. When the vibrator is set to frequency of 108 Hz. a standing wave with seventeen loops is formed. What must be the linear mass density of the cord? The acceleration of gravity is 9.8 m/s2. Answer in units of kg/m How many complete loops (if any) will result if the mass is changed to 355 kg? (Answer with -100 if no standing wave forms.) How many complete loops (if any) will result if the mass is changed to 2051.9 kg? (Answer with -100 if no standing wave forms.)

Explanation / Answer

(A) 17 loops means a length of 17(/2) ( = wavelength)

17/2 = l = 2.7

= 0.3176 m

Frequency = 198 Hz

speed of wave = wavelength x frequency

= 0.3176 x 198 = 62.89 m/s

Now, speed of wave = sqrt(T/)

(T = tension in string, = linear density)

Linear density = = T/speed^2

= mg/v^2 = (7.1 x 9.8)/(62.89^2) = 1.759 x 10^-2 Kg/m

(B) For mass = 355Kg, tension = mg = 355 x 9.8 = 3479 N

speed of wave = sqrt(tension/linear density)

= sqrt(3479/0.01759) = 444.73 m/s

For a frequency of 198 Hz, wavelength = speed/frequency

= 444.73/198 = 2.246 m

Length of rope = 2.7 m

l/ = 2.7/2.46 = 1.2

But for an integer number of loops to form, the length of the rope must be an integral multiple of /2. But as clearly illustrated above, its not. Therefore, standing waves are not formed

Ans : -100 (as requested)

(C) For mass = 2051.9 Kg, tension = mg = 2051.9 x 9.8 = 20108.62 N

speed of wave = sqrt(tension/linear density)

= sqrt(20108.62/0.01759) = 1069.198 m/s

For a frequency of 198 Hz, wavelength = speed/frequency

= 1069.198/198

= 5.399 m

5.4 m

Length of rope = 2.7 m

l/ = 2.7/5.4 = 0.5

Therefore, l = /2. Therefore a single loop is formed and the rope vibrates in its fundamental mode.

Ans : 1 loop fomed

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