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Papessa [141]
2 years ago
15

Bob creates an instrument that is able to play C4 (261.63 Hz). He does some analysis with the sound equipment and it shows that

when he plays it, he also gets the frequencies 523.26 Hz, 784.89 Hz, and 1308.15 Hz. What could be true about the instrument? It’s a stringed instrument, it’s a closed pipe, or it’s an open pipe?
Physics
1 answer:
Dmitry [639]2 years ago
4 0

Answer:

It could be a stringed instrument or an open pipe.

Explanation:

Let v be the speed of sound, y be wavelength and f be frequency.

v = yf

f= v/y; v is constant.

In a stringed instrument, the fundamental frequency note is heard when the length of the string, l = y/2; y= 2l

f′= v/2l

The second harmonic is heard when l= y

f"= v/y

...

f'''= 3v/2l

We can infer that f"= 2f'

f'''= 3f'

This is similar to the values in the question as;

523.26 =2(261.63) and so on.

Same thing happens with open pipes.

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A truck covers 40.0 m in 7.50 s while uniformly slowing down to a final velocity of 2.55 m/s. (a) Find the truck's original spee
iris [78.8K]

Answer:

(a) 8.117 (b) 0.742m/sec^2

Explanation:

We have given distance s =40 m time t=7.5 sec

Final velocity v =2.55 m/sec

From the first equation of motion v=u-at (negative sign because there is retrdation as the truck speed is slowing down )

So 2.55=u-7.5a --------------eqn 1

From the second equation of motion s=ut-\frac{1}{2}at^2 ( negative sign because there is retrdation as the truck speed is slowing down )

So 40=7.5u-0.5\times a\times 7.5^2

40=7.5u-28.125a------------------eqn 2

On solving eqn1 and eqn 2

u=8.117 m/sec and a=-0.742m/sec^2

7 0
4 years ago
(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive of a y axis with an angular wa
Vedmedyk [2.9K]

Missing data: the wave number

k=60 cm^{-1}

(a)  z = 0.003 sin (6000y-31.4 t)

For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is

z = A sin (ky-\omega t)

where

A is the amplitude of the wave

k is the wave number

\omega is the angular frequency

t is the time

In this situation:

A = 3.0 mm = 0.003 m is the amplitude

k = 60 cm^{-1} = 6000 m^{-1} is the wave number

T = 0.20 s is the period, so the angular frequency is

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s

So, the wave equation (in meters) is

z = 0.003 sin (6000y-31.4 t)

(b) 0.094 m/s

For a transverse wave, the transverse speed is equal to the derivative of the displacement of the wave, so in this case:

v_t = z' = -A \omega cos (ky-\omega t)

So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be

v_{t}_{max} =\omega A

where

\omega = 31.4 rad/s\\A = 0.003 m

Substituting,

v_{t}_{max}=(31.4)(0.003)=0.094 m/s

(c) 5.24 mm/s

The wave speed is given by

v=f \lambda

where

f is the frequency of the wave

\lambda is the wavelength

The frequency can be found from the angular frequency:

f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz

While the wavelength can be found from the wave number:

\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m

Therefore, the wave speed is

v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s

7 0
4 years ago
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