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3 years ago

9

Suppose the B string on a guitar is 24" long and vibrates with a frequency of about 247 Hz. You place your finger on the 5th fre

t, which changes the length of the string to 18". Which note do you hear when you play the string D (about 294 Hz) O E (about 329 Hz) OG (about 392 Hz)

1 answer:

pashok25 [27]3 years ago

8 0

To solve this problem it is necessary to apply the related concepts to string vibration. This concept shows the fundamental frequency of a string due to speed and length, that is,

$f = \frac{v}{2L}$

Where

v = Velocity

L = Length

Directly if the speed is maintained the frequency is inversely proportional to the Length:

$f \propto \frac{1}{L}$

Therefore the relationship between two frequencies can be described as

$\frac{f_2}{f_1}=\frac{l_1}{l_2}$

$f_2 = \frac{l_1}{l_2}(f_1)$

Our values are given as,

$l_1 = 24"\\f_1 = 247Hz\\l_2 = 18"$

Therefore the second frequency is

$f_2 = \frac{l_1}{l_2}(f_1)\\f_2 = \frac{24}{18}(247)\\f_2 = 329.33Hz$

The frequency allocation of 329Hz is note E.

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After polishing his 2-kg wrestling trophy, Mike sets it down on the ground and walks away to find more polish. Meanwhile, Julie

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1) The initial momentum of the trophy is zero

2) The initial momentum of the bowling ball is 160 kg m/s

3) The total momentum before the collision is 160 kg m/s

4) The total momentum of the system after the collision is 160 kg m/s

5) The final velocity of the trophy is 32 m/s

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1)

The momentum of an object is given by

$p=mv$

where

m is the mass of the object

v is its velocity

In this problem, the data for the trophy before the collision are:

m = 2 kg is the mass

v = 0 is its initial velocity

Therefore, the initial momentum of the trophy is

$p_1=(2)(0)=0$

2)

Using the same equation used in part 1), the initial momentum of the bowling ball is

$p=mv$

where

m is the mass of the bowling ball

v is its initial velocity

The data of the problem are

m = 8 kg is the mass

v = 20 m/s is the velocity

Substituting,

$p_2=(8)(20)=160 kg m/s$

3)

The total momentum of the system before the collision is given by the sum between the initial momentum of the trophy and the initial momentum of the bowling ball:

$p_i = p_1 + p_2$

where

$p_1$ is the initial momentum of the trophy

$p_2$ is the initial momentum of the ball

Here we have

$p_1 = 0$

$p_2 = 160 kg m/s$

Therefore, the total momentum is

$p_i = 0 + 160 = 160 kg m/s$

4)

According to the law of conservation of momentum, for an isolated system (=no external unbalanced forces acting on the system), the total momentum of the system is conserved before and after the collision:

$p_i = p_f$

where

$p_i$ is the total momentum before the collision

$p_f$ is the total momentum after the collision

If we consider the system in the problem to be isolated (i.e. no frictional forces acting on the ball or the trophy), we can therefore say that the total momentum after the collision must be equal to the total momentum before the collision: therefore,

$p_f = 160 kg m/s$

5)

We can write the total momentum after the collision as

$p_f = m_1 v_1 + m_2 v_2$

where:

$m_1 = 2 kg$ is the mass of the trophy

$v_1$ is the final velocity of the trophy

$m_2 = 8 kg$ is the mass of the bowling ball

$v_2 = 12 m/s$ is the final velocity of the ball

Since we also know the value of the final total momentum,

$p_f = 160 kg m/s$

we can solve the equation to find the velocity of the trophy:

$v_1 = \frac{p_f - m_2 v_2 }{m_1}=\frac{160-(8)(12)}{2}=32 m/s$

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