For a demonstration of standing waves a string with length 3.8 m is attached to

Question

For a demonstration of standing waves a string with length 3.8 m is attached to an oscillator operating at 65 Hz. The other end of the string passes over a pulley to a hanger where various masses can be attached to vary the tension in the string. Transverse standing waves are set up in the string with a node at the pulley and an anti-node at the oscillator. The mass per unit length of the string is 9 g/m.

(a) What must the tension be for the string to vibrate with its lowest frequency?
( ) N
(b) What tensions are needed for the three next higher frequencies?
T2=
T3=
T4=

Explanation / Answer

a) for fundamental ,

wavelength, lambda = 4L = 4 x 3.8 = 15.2 m

and speed = wavelnegth x frequency   = 15.2 x 65 = 988 m/s

and speedd, v = sqrt(T / linear density )

linear density = 9 g/m = 0.009 kg /m

988 = sqrt( T / 0.009)

T = 8785.3 N ........Ans

b) for next higher,

lambda = 4 L /3 = 4 x 3.8 / 3 = 5.07 m

and speed = wavelnegth x frequency   = 5.07 x 65 = 329.33 m/s

and speedd, v = sqrt(T / linear density )

linear density = 9 g/m = 0.009 kg /m

329.33 = sqrt( T / 0.009)

T =976.14   N ........Ans

for next highher :

lambda = L / 1.25 = 3.04 m

and speed = wavelnegth x frequency   = 3.04 x 65 = 197.6 m/s

and speedd, v = sqrt(T / linear density )

linear density = 9 g/m = 0.009 kg /m

197.6 = sqrt( T / 0.009)

T = 351.41   N ........Ans

for next higher:

lambda = L / 1.75 = 3.8/1.75 = 2.17 m

and speed = wavelnegth x frequency   = 2.17 x 65 = 141.14 m/s

and speedd, v = sqrt(T / linear density )

linear density = 9 g/m = 0.009 kg /m

141.14   = sqrt( T / 0.009)

T = 179.3 N ........Ans

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