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lesya [120]
3 years ago
8

A standing wave pattern is created on a string with mass density μ = 3.4 × 10-4 kg/m. A wave generator with frequency f = 61 Hz

is attached to one end of the string and the other end goes over a pulley and is connected to a mass (ignore the weight of the string between the pulley and mass). The distance between the generator and pulley is L = 0.62 m. Initially the 3rd harmonic wave pattern is formed.
1) What is the wavelength of the wave? m 2) What is the speed of the wave? m/s 3) What is the tension in the string? N 4) What is the mass hanging on the end of the string? kg 5)Now the hanging mass is adjusted to create the 2nd harmonic. The frequency is held fixed at f = 61 Hz.What is the wavelength of the wave? m 6) What is the speed of the wave? m/s 7) What is the tension in the string? N 8) What is the mass hanging on the end of the string? kg 9) Keeping the frequency fixed at f = 61 Hz, what is the maximum mass that can be used to still create a coherent standing wave pattern?
Physics
1 answer:
uranmaximum [27]3 years ago
5 0

Answer:

1) λ = 0.413 m , 2)v = 25,213 m / s , 3)  T = 0.216 N , 4) m = 22.04 10-3 kg

Explanation:

1) The resonance occurs when the traveling wave bounces at the ends and the two waves are added, the ends as they are fixed have a node, the wavelength and the length of the string are related

         λ = 2L / n               n = 1, 2, 3 ...

In this case L = 0.62 m and n = 3

Let's calculate

        λ = 2 0.62 / 3

        λ = 0.413 m

2) the velocity related to wavelength and frequency

      v =  λ f

      v = 0.413 61

      v = 25,213 m / s

3) let's use the equation

     v = √T /μ

     T = v² μ

     T = 25,213² 3.4 10⁻⁴

     T = 0.216 N

4) the rope tension is proportional to the hanging weight

      T-W = 0

     T = W

    W = m g

    m = W / g

    m = 0.216 / 9.8

    m = 22.04 10-3 kg

5) n = 2

     λ = 2 0.62 / 2

     λ = 0.62 m

6) v =  λ f

     v = 0.62 61

     v = 37.82 m / s

7) T = v² μ

   T = 37.82² 3.4 10⁻⁴

   T = 0.486 N

8) m = W / g

   m = 0.486 / 9.8

   m = 49.62 10⁻³ kg

9) n = 1

    λ = 2 0.62

    λ = 1.24 m

    v = 1.24 61

    v = 75.64 m / s

    T = v² miu

    T = 75.64² 3.4 10⁻⁴

    T = 2.572 10⁻² N

    m = 2.572 10⁻² / 9.8

    m = 262.4 10⁻³ kg

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⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a)
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Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

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g = 9.8m/s^2

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Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

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(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

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