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Marat540 [252]
3 years ago
15

A large ant is standing on the middle of a circus tightrope that is stretched with tension Ts. The rope has mass per unit length

μ. Wanting to shake the ant off the rope, a tightrope walker moves her foot up and down near the end of the tightrope, generating a sinusoidal transverse wave of wavelength λ and amplitude A. Assume that the magnitude of the acceleration due to gravity is g.What is the minimum wave amplitude Amin such that the ant will become momentarily "weightless" at some point as the wave passes underneath it? Assume that the mass of the ant is too small to have any effect on the wave propagation.Express the minimum wave amplitude in terms of Ts, μ, λ, and g.
Physics
1 answer:
natta225 [31]3 years ago
8 0

Answer:

A = g (λ / 2π)² μ / Ts

Explanation:

The ant becomes "weightless" when its acceleration is equal to gravity.  The motion of the ant is a sinusoidal wave:

y = A sin(ωt)

By taking the derivative twice, we can find the acceleration of the ant:

y' = Aω cos(ωt)

y" = -Aω² sin(ωt)

The maximum acceleration occurs when sine is 1.  We want this to happen at a = -g.

-g = -Aω² (1)

A = g / ω²

Angular frequency is 2π times the normal frequency:

ω = 2πf

A = g / (2πf)²

Frequency is velocity divided by wavelength:

f = v / λ

A = g / (2πv / λ)²

A = g (λ / 2π)² / v²

Velocity of a wave in a string with tension Ts and linear density μ is:

v = √(Ts / μ)

Therefore:

v² = Ts / μ

Plugging in:

A = g (λ / 2π)² / (Ts / μ)

A = g (λ / 2π)² μ / Ts

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Two loudspeakers in a 20c room emit 686 hz sound waves along the x-axis.a. if the speakers are in phase, what is the smallest di
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<u>Answer :</u>

(a) d = 0.25 m

(b) d = 0.5 m

<u>Explanation :</u>

It is given that,

Frequency of sound waves, f = 686 Hz

Speed of sound wave at 20^0\ C is, v = 343 m/s

(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.

d=\dfrac{\lambda}{2}........(1)

Velocity of sound wave is given by :

v=f\times \lambda

d=\dfrac{v}{2f}

d=\dfrac{343\ m/s}{2\times 686\ Hz}

d=0.25\ m

Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.

(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.

d=n\lambda  ( n = integers )

Let n = 1

So, d=\dfrac{v}{f}

d=\dfrac{343\ m/s}{686\ Hz}

d=0.5\ m

Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.

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Explanation:

It is given that,

Velocity in East, v_1=4\ m/s

Velocity in North, v_2=3\ m/s

(a) The resultant velocity is given by :

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(b) The width of the river is, d = 80 m

Let t is the time taken by the boat to travel shore to shore. So,

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(c) Let x is the distance covered by the boat to reach the opposite shore. So,

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Hence, this is the required solution.

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