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4 years ago

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A wave pulse travels along a string at a speed of 230 cm/s . Note that parts a - d are independent and refer to changes made to

the original string.(a) What will be the speed if the string's tension is doubled?(b) What will be the speed if the string's mass is quadrupled (but its length is unchanged)?(c) What will be the speed if the string's length is quadrupled (but its mass is unchanged)?

1 answer:

Oksanka [162]4 years ago

3 0

a) The speed of the wave will increase by a factor $\sqrt{2}$ .

b) The speed of the wave will halve

c) The speed of the wave will double

Explanation:

a)

The speed of a standing wave on a string is given by

$v=\sqrt{\frac{T}{m/L}}$

where

T is the tension in the string

m is the mass of the string

L is the length of the string

In this part of the problem, the tension in the string is doubled, so that the new tension is

T' = 2T

Substituting into the equation, we find the new speed of the wave in the string:

$v'=\sqrt{\frac{T'}{m/L}}=\sqrt{\frac{2T}{m/L}}=\sqrt{2}\sqrt{\frac{T}{m/L}}=\sqrt{2}v$

So, the speed will increase by a factor $\sqrt{2}$ .

b)

We can solve also this part by referring to the formula

$v=\sqrt{\frac{T}{m/L}}$

where

T is the tension

m is the mass

L is the length

In this case, the string mass is quadrupled, so the new mass is:

m' = 4m

Substituting into the equation, we find what happens to the speed of the wave:

$v'=\sqrt{\frac{T}{m'/L}}=\sqrt{\frac{T}{4m/L}}=\frac{1}{\sqrt{4}}\sqrt{\frac{T}{m/L}}=\frac{1}{2}v$

So, the speed of the wave will halve.

c)

Again, we can solve this part by referring to the same equation

$v=\sqrt{\frac{T}{m/L}}$

where

T is the tension

m is the mass

L is the length

In this case, the length of the string is quadrupled, so the new length is:

L' = 4L

Substituting into the equation, we find that the new speed is:

$v'=\sqrt{\frac{T}{m/L'}}=\sqrt{\frac{T}{m/(4L)}}=\sqrt{4}\sqrt{\frac{T}{m/L}}=2v$

So, the speed of the wave will double.

Learn more about waves:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

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