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sleet_krkn [62]
3 years ago
11

Calculate the frequency of the 3rd normal mode of a guitar string of length 40.0cm and mass 0.5g when stretched with a tension o

f 80N.
Physics
1 answer:
Alexandra [31]3 years ago
7 0

Answer:

Explanation:

For third normal mode of vibration

l = \frac{3\lambda}{2}  , λ is wavelength , l is length of string .

.4 = \frac{3\lambda}{2}

λ = .267 m

velocity = \sqrt{\frac{T}{m} }

T is tension and m is mass unit length

m = .5 x 10⁻³ / 40 x 10⁻²

= .00125 kg / m

Putting the values

velocity = \sqrt{\frac{80}{.00125} }

= 253 m /s

frequency

= velocity / λ

= 253 / .267

= 947.5 Hz .

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Hiran is standing beside the road when he hears a bird flying away from hip and chirping. The bird’s chirp has a frequency of 18
Murrr4er [49]

The frequency of bird chirping hear by hiran will be 1.77 kHz.

<u>Explanation:</u>

As per Doppler effect, the observer will feel a decrease in the frequency of the receiving signal if the source is moving away from the observer. So the shifted frequency is obtained using the below equation:

f'=\frac{c}{c+v_{s} }f

Here , c is the speed of sound, Vs is the velocity of source with which it is moving away. f is the original frequency of source and f' is the frequency shift heard by the observer.

As here, f = 1800 Hz, Vs= 6 m/s and c = 343 m/s, then

f'=\frac{343}{343+6} \times 1800=1.77\ kHz

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A transverse, wave travelling on a chord is represented by D=0.22sin (5.6x+34t) where D and x are inmeters and t is in seconds.
ArbitrLikvidat [17]

Answer:

a) λ = 1.12 m

b) f = 5.41 Hz

c) v = 154.54 m/s

d) A = 0.22m

e)

v_D_{max}=7.48\frac{m}{s}\\\\v_D_{min}=-7.48\frac{m}{s}\\\\

Explanation:

You have the following equation for a wave traveling on a cord:

D=0.22sin(5.6x+34t)     (1)

The general expression for a wave is given by:

D=Asin(kx-\omega t)    (2)

By comparing the equation (1) and (2) you have:

A: amplitude of the wave = 0.22m

k: wave number = 5.6 m^-1

w: angular velocity = 34 rad/s

a) The wavelength is given by substitution in the following expression:

\lambda=\frac{2\pi}{k}=\frac{2\pi}{5.6m^{-1}}=1.12m

b) The frequency is:

f=\frac{\omega}{2\pi}=\frac{34s^{-1}}{2\pi}=5.41Hz

c) The velocity of the wave is:

v=\frac{\omega}{k}=\frac{34s^{-1}}{0.22m^{-1}}=154.54\frac{m}{s}

d) The amplitude is 0.22m

e) To calculate the maximum and minimum speed of the particles you obtain the derivative of  the equation of the wave, in time:

v_D=\frac{dD}{dt}=(0.22)(34)cos(5.6x+34t)\\\\v_D=7.48cos(5.6x+34t)

cos function has a minimum value -1 and maximum +1. Then, you obtain for maximum and minimum velocity:

v_D_{max}=7.48\frac{m}{s}\\\\v_D_{min}=-7.48\frac{m}{s}\\\\

6 0
3 years ago
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