A string is stretched between a fixed support and a pulley a distance 103 cm apa

Question

A string is stretched between a fixed support and a pulley a distance 103 cm apart. The tension on the string is controlled by a weight hanging from the string below the pulley. An electomechnical vibrator is used to vibrate the string at 120 Hz. When the weight is adjusted to 285 g, the string is found to oscillate in its third standing wave mode (three loops). What is the speed of wave propogation on the string in this mode?
What mass must be suspended from the string to produce instead the second standing wave mode (two loops)?

Explanation / Answer

given

length = 103 cm

frequency = 120 hz

weight = 285 gm

120 hz = 3 * fundamental frequency

fundamental frequency = 40 Hz

fundamental frequency = v / 0.206

40 * 0.206= v

v = 8.24 m/s

v = sqrt(T / m/L)

8.24 = sqrt(0.285 * 9.8 / x)

x = 0.0411

velocity if second standing wave

fundamental frequency = 120 / 2

fundamental frequency = 60 Hz

60 = v / 0.206

v = 12.36 m/s

12.36 = sqrt( m * 9.8 / 0.0411 )

m = 0.6407 kg

mass must be suspended from the string to produce instead the second standing wave mode = 0.6407 kg or 640.7 g

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