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

The fundamental frequency of the vibrating cord is approximately 14.18 Hz.

Step-by-step explanation:

To determine the fundamental frequency of the vibrating cord, we can use the equation for the fundamental frequency of a vibrating string:

f1 = 1/2L * sqrt(T / μ)

Where f1 is the fundamental frequency, L is the total length of the cord, T is the tension in the cord, and μ is the linear density of the cord.

Plugging in the given values, we have: f1 = 1/2 * 7.00 * sqrt(90 / 0.007)

Simplifying this equation gives us the fundamental frequency of the cord: f1 ≈ 14.18 Hz

Answer 2

frequency = 19.56 Hz

Step-by-step 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 =
`(v)/(2d)` ...........1

so here d is given = 4

and v =
`\sqrt{(T)/(\mu) }` ..........2

tension T = Mg = 2.50 × 9.8 = 24.5 N

and μ =
`(m)/(l)` =
`(7*10^(-3) )/(7)` =
`10^(-3)` kg/m

so from equation 2

v =
`\sqrt{(24.5)/(10^(-3)) }`

v = 156.52

and from equation 1

frequency =
`(v)/(2d)`

frequency =
`(156.52)/(2(4))`

frequency = 19.56 Hz