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3 years ago

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A horizontal wire is stretched with a tension of 98.4 N, and the speed of transverse waves for the wire is 406 m/s. What must th

e amplitude of a traveling wave of frequency 60.5 Hz be for the average power carried by the wave to be 0.391 W? Give an answer in mm. Pay attention to the number of significant figures.

1 answer:

Anika [276]3 years ago

4 0

Answer:

Amplitude = 4.725 mm

Explanation:

We have got the following values :

T = 98.4 N

T is the wire tension

v = 406 m/s

v is the transverse waves speed

f = 60.5 Hz

f is the frequency

ΔU = 0.391 W

ΔU is the average power carried by the wave

If μ is the mass per unit length of the wire ⇒ μ = $\frac{T}{v^{2}}$

If ω is the angular frequency of the wire ⇒ ω = 2π.f

So ω = 2π(60.5 Hz) and ω units are $s^{-1}$

The equation that relates all is :

ΔU = (1/2).(T/v^2).(ω^2).(A^2).v

Where A is the wave amplitude

$0.391 W=\frac{1}{2}.\frac{98.4 N}{(406\frac{m}{s})^{2}}.(2.pi.(60.5Hz))^{2}.A^{2}.406\frac{m}{s}$

This will give us A in meters

$A=4.7253(10^{-3})m\\A=4.725 mm$

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b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

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a)

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

$F_{net}=ma$

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In the first 0.890 s of motion, we have

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$a=9.50 m/s^2$ (acceleration)

So, the net force is

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b)

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

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u is the initial speed

a is the acceleration

t is the time

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$a=9.50 m/s^2$ (acceleration)

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First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

$W=K_f - K_i = \frac{1}{2}mv^2-0$

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

$K_i = 0 J$ is the initial kinetic energy

So the work done is

$W=\frac{1}{2}(94.0)(8.5)^2=3396 J$

The power expended is given by

$P=\frac{W}{t}$

where

t = 0.890 s is the time elapsed

Substituting,

$P=\frac{3396}{0.890}=3816 W$

d)

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

$F_1=893 N$

We know that the average force for the whole race is

$F_{avg}=820 N$

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$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}$

And solving for $F_2$ , we find the average force exerted by Bolt on the ground during the second phase:

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N$

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$F_{net}=F_2-D$ (1)

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