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slamgirl [31]
3 years ago
7

A guitar string with a linear density of 2.0 g/m is stretched between supports that are 60 cm apart. The string is observed to f

orm a standing wave with three antinodes when driven at a frequency of 420 Hz.
Part A
What is the frequency of the fifth harmonic of this string?
Physics
1 answer:
Butoxors [25]3 years ago
3 0

<h2>The frequency of driver is 700 Hz</h2>

Explanation:

The frequency of wave in a string is given by the relation

n = \frac{p}{2l} \sqrt{\frac{T}{m} }

here n is the frequency

p is the number of antinodes and l is the length of string .

T is the tension in string and m is the mass per unit length

Thus 420 =\frac{3}{120} \sqrt{\frac{T}{2} }            I

Now if there is 5 antinodes , the value of p = 5

Thus n = \frac{5}{120} \sqrt{\frac{T}{2} }          II

Dividing II by I , we have

n/420 = 5/3

or n = 5/3 x 420 = 700 Hz

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Well she drove 30 more miles to the east than the west but I don’t understand what u are asking
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3 years ago
1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.
marta [7]

(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

<h3>Surface area being vibrated</h3>

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

<h3>Intensity of the sound</h3>

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

<h3>Relative intensity of the sound</h3>

B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB

<h3>Speed of sound at the given temperature</h3>

v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s

<h3>Frequency of the sound</h3>

The frequency of the sound heard is determined by applying Doppler effect.

f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )

where;

  • -v₀ is velocity of the observer moving away from the source
  • -vs is the velocity of the source moving towards the observer
  • fs is the source frequency
  • fo is the observed frequency
  • v is speed of sound

f_0 = f_s(\frac{v-v_0}{v- v_s} )

f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz

Learn more about intensity of sound here: brainly.com/question/17062836

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2 years ago
The uniform rods AB and BC weigh 24 ky and kg, respectively,and the small wheel at C is of negligible weight. If the wheel ismov
victus00 [196]

The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

T_{1}  + V_{1} = T_{2}  + V_{2}

h_{ab}  = h_{bc}  = 0.18m

potential energy at position 1,

V1 = m_{ab} gh_{ab}  + m_{bc} gh_{bc}

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} +  \frac{1}{2} m_{bc} v^{2}_{G} +  \frac{1}{2} lw^{2}_{bc}

l_{ab} =\frac{m_{ab} l^{2}_{ab}  }{3}

= 1/3 *4 * (0.36)²

=0.10368kg m²

l =\frac{m_{bc} l^{2}_{bc}  }{12}

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

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5 0
1 year ago
What affect does a doubling of the net force have upon the acceleration of the object?
balandron [24]
F = ma
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3 years ago
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Hooke's law describes a certain light spring of unstretched length 34.8 cm. When one end is attached to the top of a doorframe a
DaniilM [7]

Answer:

a) k = 1343.6\,\frac{N}{m}, b) l = 0.501\,m\,(50.1\,cm)

Explanation:

a) The Hooke's law states that spring force is directly proportional to change in length. That is to say:

F \propto \Delta l

In this case, the force is equal to the weight of the object:

F = m\cdot g

F = (8.22\,kg)\cdot (9.807\,\frac{m}{s^{2}} )

F = 80.614\,N

The spring constant is:

k = \frac{F}{\Delta l}

k = \frac{80.614\,N}{0.408\,m-0.348\,m}

k = 1343.6\,\frac{N}{m}

b) The length of the spring is:

F = k\cdot (l-l_{o})

l = l_{o} + \frac{F}{k}

l=0.348\,m+\frac{205\,N}{1343.6\,\frac{N}{m} }

l = 0.501\,m\,(50.1\,cm)

6 0
3 years ago
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