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elena-14-01-66 [18.8K]
2 years ago
8

You have a string with a mass of 0.0123 kg.0.0123 kg. You stretch the string with a force of 8.37 N,8.37 N, giving it a length o

f 1.81 m.1.81 m. Then, you vibrate the string transversely at precisely the frequency that corresponds to its fourth normal mode; that is, at its fourth harmonic. What is the wavelength ????4λ4 of the standing wave you create in the string? What is the frequency f4?
Physics
1 answer:
Elza [17]2 years ago
4 0

Answer:

v= 35.09m/s

f= 38.77Hz

Explanation:

Length 'L' = 1.81 m

Now, First off the fourth harmonic of a string with two nodes is given by:

L=2λ

1.81=2λ

λ= 1.81/2 =>0.905 m

Wave speed on a strecthed string is:

v=√(Ft/(m/l))

v= √(8.37)/(0.0123/1.81)

v= 35.09m/s

Next is to determine frequency i,e

v=λf or f =v/λ

f= 35.09/0.905

f= 38.77Hz

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An 80 g, 40 cm long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 15 g ball of clay trav
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Answer:

θ  = 12.95º

Explanation:

For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height

Let's start by finding the speed of the bar plus clay ball system, using amount of momentum

The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)

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Final after the crash before starting the movement

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With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy

Lower

    Em₀ = K = ½ (m + M) v²

Higher

    E_{mf} = U = (m + M) g y

   Em₀ = E_{mf}

   ½ (m + M) v² = (m + M) g y

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Let's look for the angle the height from the pivot point is

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The distance that went up is

     y = L - L cos θ

     cos θ  = (L-y) / L

     θ  = cos⁻¹ (L-y) / L

     θ  = cos⁻¹-1 ((0.20 - 0.00509) /0.20)

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