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Anna35 [415]
3 years ago
13

Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co

mpleted in 22.7 s. When this motion is viewed as a projection of circular motion, what are the radius, r, and angular velocity, ? , of the circular motion?
Physics
1 answer:
aivan3 [116]3 years ago
4 0

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m

The radius is 1.195 m

Time period would be given by

T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s

Time period of the motion is 2.8375 s

Angular speed is given by

\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s

The angular speed of the motion is 2.21433 rad/s

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Here's the formula for the distance covered by an accelerating body in some amount of time ' T '. This formula is incredibly simple but incredibly useful. It pops up so often in Physics that you really should memorize it:

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This question gives us the acceleration and the distance, and we want to find the time.

(9,000 m) = (1/2) (20 m/s²) (time²)

(9,000 m) = (10 m/s²) (time²)

Divide each side by 10 m/s²:

(9,000 m) / (10 m/s²) = (time²)

900 s² = time²

Square root each side:

<em>T = 30 seconds</em>

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The current in a toaster is 20 amps, and its voltage is 200 volts. What is the resistance in the toaster?
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A wind instrument that produces only odd-numbered standing wave modes has what configuration of its ends?
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Open- closed

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3 years ago
One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe
Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

\mu = \frac{m}{l}

\mu = \frac{8.79*10^{-3}}{70*10^{-2}}

\mu = 0.01255kg/m

The expression for the wavelength of the standing wave for the second overtone is

\lambda = \frac{2}{3} l

Replacing we have

\lambda = \frac{2}{3} (70*10^{-2})

\lambda = 0.466m

The frequency of the sound wave is

f_s = \frac{v}{\lambda_s}

f_s = \frac{344}{0.768}

f_s = 448Hz

Now the velocity of the wave would be

v = f_s \lambda

v = (448)(0.466)

v = 208.768m/s

The expression that relates the velocity of the wave, tension on the string and linear mass density is

v = \sqrt{\frac{T}{\mu}}

v^2 = \frac{T}{\mu}

T= \mu v^2

T = (0.01255kg/m)(208.768m/s)^2

T = 547N

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the n^{th} harmonic frequency is

f_n = nf_1

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

n=3

Then,

f_3 = 3f_1

Rearranging to find the fundamental frequency

f_1 = \frac{f_3}{3}

f_1 = \frac{448Hz}{3}

f_1 = 149.9Hz

7 0
3 years ago
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