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Lemur [1.5K]
3 years ago
7

One of the 63.5-cm-long strings of an ordinaryguitar is tuned to produce the note {\rm B_3}(frequency 245 Hz) when vibrating ini

ts fundamental mode.
A.Find the speed of transverse waves on this string.B.If the tension in this string is increased by1.0%, what will be the new fundamental frequency ofthe string?C.If the speed of sound in the surrounding air is 344m/s, find the frequency of the sound wave produced in theair by the vibration of the {\rm B_3} string.D.If the speed of sound in the surrounding air is 344m/s, find the wavelength of the sound wave produced in theair by the vibration of the {\rm B_3} string.
Physics
1 answer:
lubasha [3.4K]3 years ago
3 0

Answer:

a,V=311.15m/s

b.246Hz

c.245Hz

d. 1.4m

Explanation:

One of the 63.5-cm-long strings of an ordinaryguitar is tuned to produce the note {\rm B_3}(frequency 245 Hz) when vibrating inits fundamental mode.

A.Find the speed of transverse waves on this string.B.If the tension in this string is increased by1.0%, what will be the new fundamental frequency ofthe string?C.If the speed of sound in the surrounding air is 344m/s, find the frequency of the sound wave produced in theair by the vibration of the {\rm B_3} string.D.If the speed of sound in the surrounding air is 344m/s, find the wavelength of the sound wave produced in theair by the vibration of the {\rm B_3} string

NOTE

speed is distance by the wave per time

frequency is the number of oscillation the wave front makes in one seconds

the wave speed is given by

v=\sqrt{F/U}

recall also that the wave speed is v=f lambda

for  a standing wave , we know the fundamental frequency of a string is

f1=v/2L

L=length of the string

f1=245Hz

V=?

L=0.635m

V=245*2*0.635

V=311.15m/s

b. tension in the string is increased by 1%

F2=F+1%

f2=101F%

substituting for F2 into this equation

v=\sqrt{F/U}

v2=\sqrt1.01F/U}

v2=\sqrt{1.01} v

v2=1.01^0.5*311.15m/s

v2=312.7m/s

for the new fundamental frequency we have

f2=312.7/2*0.635

f2=246Hz

c. the frequency of the sound wave equal the frequency of the string that created it

c. fs=245Hz

d. speed of sound in air344m/s

v=344m/s

v=f*lambda

lambda is the wavelength

344=245*lambda

344/245=1.40m

wavelength of string B3 is 1.4m

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Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor
fenix001 [56]

Answer:

a) F_{r}= -583.72MN i + 183.47MN j + 6.05GN k

b) E=3.04 \frac{GN}{C}

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}

which yields:

r_{AP}^{2}= 43 m^{2}

(I will assume the positions are in meters)

Next, we can make use of the force formula:

F=k_{e}\frac{q_{1}q_{2}}{r^{2}}

so we substitute the values:

F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}

which yields:

F_{AP}=418.14 MN

Now we can find its components:

F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i

F_{APx}=191.30 MNi

F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j

F_{APy}=191.30MN j

F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k

F_{APz}=318.83 MN k

And we can now write them together for the first force, so we get:

F_{AP}=(191.30i+191.30j+318.83k)MN

We continue with the next force. The procedure is the same so we get:

r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}

which yields:

r_{BP}^{2}= 33 m^{2}

Next, we can make use of the force formula:

F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}

which yields:

F_{BP}=2.18 GN

Now we can find its components:

F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i

F_{BPx}=758.98 MNi

F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j

F_{BPy}=758.98MN j

F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k

F_{BPz}=1.897 GN k

And we can now write them together for the second, so we get:

F_{BP}=(758.98i + 758.98j + 1897k)MN

We continue with the next force. The procedure is the same so we get:

r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}

which yields:

r_{CP}^{2}= 30 m^{2}

Next, we can make use of the force formula:

F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}

which yields:

F_{CP}=4.20 GN

Now we can find its components:

F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i

F_{CPx}=-1.534 GNi

F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j

F_{CPy}=-766.81 MN j

F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k

F_{CPz}=3.83 GN k

And we can now write them together for the third force, so we get:

F_{CP}=(-1.534i - 0.76681j +3.83k)GN

So in order to find the resultant force, we need to add the forces together:

F_{r}=F_{AP}+F_{BP}+F_{CP}

so we get:

F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN

So when adding the problem together we get that:

F_{r}=(-0.583.72i + 0.18347j +6.05k)GN

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}

which yields:

F_{r}=6.08 GN

and now we take the formula for the electric field which is:

E=\frac{F_{r}}{q}

so we go ahead and substitute:

E=\frac{6.08GN}{2C}

E=3.04\frac{GN}{C}

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