Question

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?

Answer 1

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 = $\frac{v}{2d}$   ...........1

so here d is given = 4

and v = $\sqrt{\frac{T}{\mu} }$  ..........2

tension T = Mg = 2.50 × 9.8 = 24.5 N

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

so from equation 2

v = $\sqrt{\frac{24.5}{10^{-3}} }$

v = 156.52

and from equation 1

frequency = $\frac{v}{2d}$

frequency = $\frac{156.52}{2(4)}$

frequency = 19.56 Hz