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The total length of the cord is L = 7.00 m, the mass of the cord is m = 7.00 g, the mass of the hanging object is M = 2.50 kg, and the pulley is a fixed a distance d = 4.00 m from the wall. You pluck the cord between the wall and the pulley and it starts to vibrate. What is the fundamental frequency (in Hz) of its vibration?

Respuesta :

Answer:

frequency = 19.56 Hz

Explanation:

given data

length L = 7 m

mass m = 7 g

mass M = 2.50 kg

distance d = 4 m

to find out

fundamental frequency

solution

we know here frequency formula is

frequency = [tex]\frac{v}{2d}[/tex]   ...........1

so here d is given = 4

and v = [tex]\sqrt{\frac{T}{\mu} }[/tex]  ..........2

tension T = Mg = 2.50 × 9.8 = 24.5 N

and μ = [tex]\frac{m}{l}[/tex] =  [tex]\frac{7*10^{-3} }{7}[/tex] = [tex]10^{-3}[/tex] kg/m

so from equation 2

v = [tex]\sqrt{\frac{24.5}{10^{-3}} }[/tex]

v = 156.52

and from equation 1

frequency = [tex]\frac{v}{2d}[/tex]

frequency = [tex]\frac{156.52}{2(4)}[/tex]

frequency = 19.56 Hz

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