A string of total mass 5.00×10 -2 kg is stretched to a length of 3.7m by applyin

Question

A string of total mass 5.00×10-2kg is stretched to a length of 3.7m by applying a tension of 12.10N. You flick the end of the string, sending a pulse to the far end. It reflects back to your hand. How fast is the pulse moving?

How much time does it take for the pulse to make the round trip (from your hand to the other end and back again)?

The mass of a string is 4.85E-3 kg, and it is stretched so that the tension in it is 190 N. A transverse wave traveling on this string has a frequency of 270 Hz and a wavelength of 0.521 m. What is the length of the string?

The tension in a string is 15.9 N, and its linear density is 0.860 kg/m. A wave on the string travels toward the -x direction; it has an amplitude of 3.44 cm and a frequency of 10.3 Hz. What is the speed of the wave?

Explanation / Answer

Velocity, v = SQRT[T/ `]
T is the tension in the string, T = 12.10 N
is the mass per unit length, = (5 x 10-2) / 3.7
= 0.0135 kg/m

Substituting,
v = SQRT [12.1 / (0.0135)]
= SQRT[895.4]
v = 29.92 m/s

b)
Total length traveled by the wave = 2L
                                                = 2 x 3.7 = 7.4 m
Velocity, v = 29.92 m/s
Time taken = 2L/v
                = 7.4 / 29.92
                = 0.247s.

Q-2)

Wave velocity = f = 270Hz x 0.521m

v = 140.67 m/s

Wave velocity on taut string v = (T.L/m)

                     where, T= tension, L=length, m=mass
v² = T.L/m
L = v² m/T

L = 140.67² *(4.85*10*10-3kg) / 190 N

L = 0.505 m

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