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Grade: 12

                        

The length of the wire shown in figure between the pulleys is 1.5 m and its mass is 12.0 g. Find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.

3 years ago

Answers : (1)

Arun
24737 Points
							
Tension T= mg That is T= 9×9.8 = 88.2 N . mass per unit length of the string can be calculated, this is equal to 0.8 × 10^ -2 we know velocity of wave = √{tension/mass per unit length}So wave velocity is 105m/sIts given that there are 2 loops and the entire wire is vibrating except the mid point so it becomes a node. And 2loops , one up second down on either side of mid point , A wavelength. wavelength = length of wire 1.5 mWe know wavelength , velocity So frequency can be calculated by v=∆f where ∆ = wavelength Frequency is 70 Hz
3 years ago
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