The longer the wavelength, the lower the frequency, and the lower the energy. Lesson 4.07 True False

Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note

Sophie [7]

(a) 2 Hz

The frequency of the nth-harmonic is given by

$f_n = n f_1$

where

$f_1$ is the fundamental frequency

Therefore, the frequency of the third harmonic of the A ( $f_1 = 440 Hz$ ) is

$f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz$

while the frequency of the second harmonic of the E ( $f_1 = 659 Hz$ ) is

$f_2 = 2 \cdot f_1 = 2 \cdot 659 Hz =1318 Hz$

So the frequency difference is

$\Delta f = 1320 Hz - 1318 Hz = 2 Hz$

(b) 2 Hz

The beat frequency between two harmonics of different frequencies f, f' is given by

$f_B = |f'-f|$

In this case, when the strings are properly tuned, we have

- Frequency of the 3rd harmonic of A-note: 1320 Hz

- Frequency of the 2nd harmonic of E-note: 1318 Hz

So, the beat frequency should be (if the strings are properly tuned)

$f_B = |1320 Hz - 1318 Hz|=2 Hz$

(c) 1324 Hz

The fundamental frequency on a string is proportional to the square root of the tension in the string:

$f_1 \propto \sqrt{T}$

this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.

In this situation, the beat frequency is 4 Hz (four beats per second):

$f_B = 4 Hz$

And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,

$f_B = |f_3-f_2|$

and $f_3 = 1320 Hz$ , we have two possible solutions for $f_2$ :

$f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz$

However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be

1324 Hz

8 0

3 years ago

Based on our study of electromagnets, what would be the best thing to do to a generator in order to produce more electricity?

Keith_Richards [23]

Answer:

A

Explanation:

just took a test on ed sorry if wrong I got it right

4 0

3 years ago

Read 2 more answers

A rotating object starts from rest at t = 0 s and has a constant angular acceleration. At a time of t = 2.5 s the object has an

Evgesh-ka [11]

Answer:

52 rad

Explanation:

Using

Ф = ω't +1/2αt²................... Equation 1

Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.

Since the object states from rest, ω' = 0 rad/s.

Therefore,

Ф = 1/2αt²................ Equation 2

make α the subject of the equation

α = 2Ф/t².................. Equation 3

Given: Ф = 13 rad, t = 2.5 s

Substitute into equation 3

α = 2(13)/2.5²

α = 26/2.5

α = 4.16 rad/s².

using equation 2,

Ф = 1/2αt²

Given: t = 5 s, α = 4.16 rad/s²

Substitute into equation 2

Ф = 1/2(4.16)(5²)

Ф = 52 rad.

6 0

3 years ago

Read 2 more answers

Work done in taking charge from one point of a conductor to is another point is called

Yuliya22 [10]

Answer:

⁸

Explanation:

electric potential

I think so

6 0

3 years ago

Which of the following is not possible? A. Gas flow equals pressure gradient over resistance. B. Resistance equals pressure grad

kaheart [24]

Answer:

C. Pressure gradient equals gas flow over resistance.

Explanation:

As we know that pressure gradient is the driving force for the gas to flow from one point to other point

And we know that the flow rate is directly proportional to the driving force and it inversely depends on the resistance to flow

so we can say

Flow Rate = $\frac{Driving \: force}{Resistance}$

Flow Rate = $\frac{Pressure \: Gradient}{Resistance}$

so we can say that correct statements are as below

A. Gas flow equals pressure gradient over resistance.

B. Resistance equals pressure gradient over gas flow.

D. The amount of gas flowing in and out of the alveoli is directly proportional to the difference in pressure or pressure gradient between the external atmosphere and the alveoli.

5 0

3 years ago

The longer the wavelength, the lower the frequency, and the lower the energy. Lesson 4.07 True False​

The longer the wavelength, the lower the frequency, and the lower the energy. Lesson 4.07 True False