An 385 kg mass is hung from the end of a string with linear density 2.4 g/m arou

Question

An 385 kg mass is hung from the end of a string with linear density 2.4 g/m around a small frictionless peg. A vibrator is attached at a point near the end of the string. For some values of the vibrator’s frequency and
this mass the string resonates with visible standing waves. The vibrating length of the string is 2 m.

An 385 kg mass is hung from the end of a string with linear density 2.4 g/m around a small frictionless peg. A vibrator is attached at a point near the end of the string. For some values of the vibrator's frequency and this mass the string resonates with visible standing waves. The vibrating length of the string is 2 m. What is the frequency that will produce the standing wave shown? The acceleration of gravity is 9.8 m/s2 . Answer in units of Hz.

Explanation / Answer

The mass hung to the string m = 385 kg then the tension in the string T = mg = (385 kg) (9.8 m/s^2)                                                        = 3773 N The linear mass density = 2.4*10^-3 kg/m The length of the string L = 2m The nymber of loops from the figure n = 7 Then the speed of the wave           v = T/ = (3773N)/(2.4*10^-3)                          = 1253.8 m/s The frequency of the standing wave           f = n v/2L              = (7) (1253.8m/s) / 2(2m)             = 2194.15 Hz

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