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Jobisdone [24]
2 years ago
8

A viola string with a fundamental frequency of D4 (293 Hz) is generally tuned using a tension of 49.0 N. However, just before a

concert, the string breaks and only a G3 string (196 Hz) is available. The strings are the same length and are usually tuned with the same tension. What does the tension need to be on the replacement string to bring it up to the D4 frequency
Physics
1 answer:
Alborosie2 years ago
8 0

Answer:

Explanation:

For fundamental frequency in a vibrating string , the formula is

n = 1 / 2L  x  √ ( T /m₁ )

n is frequency , L is length , T is tension and m₁ is mass per unit length .

For first string ,

293 =  1 / 2L  x  √ ( 49 N  /m₁ )

For second string , let mass per unit length be m₂ .

196 =  1 / 2L  x  √ ( 49 N  /m₂ ) ------ ( 1 )

To bring its frequency back to previous one let tension be T

293  =  1 / 2L  x  √ ( T  /m₂ ) ------- ( 2 )

Dividing

293 / 196 = √ ( T  /49  )

1.4948 = √ ( T  /49  )

2.2344 = T  /49

T = 109.48 N .

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U = 3, 9 , v = 4, 2 (a) find the projection of u onto v. (b) find the vector component of u orthogonal to v.
ExtremeBDS [4]

Answer:

(a) At U = 3, 9 , v = 4, 2, the projection of u onto v is w1=<2,8>

(b)At U = 3, 9 , v = 4, 2,  the vector component of u orthogonal to v is w2 = <4,-1>

Explanation:

A

The projection of u onto v is given by:

w1= projvu = (u⋅v||v||2)v

Given that  u= <6,7> and v=<1,4>, we can find the projection of u onto v as shown below:

w1= projvu = (u⋅v||v||2v=(<6,7>⋅<1,4><1,4>⋅<1,)

=(6⋅1+7⋅41⋅1+4⋅4)<1,4>

=3417<1,4>

=<2,8>

Part B

The vector component of u orthogonal to v is given by:

Using the given vectors and the projection found in part (a), we can find the vector component of u orthogonal to v as shown below:

w2=u−projvu

=<6,7≻<2,8>

=<(6−2),(7−8)>

=<4,−1>

To learn more about vector component, click brainly.com/question/17016695

#SPJ4

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1 year ago
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