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SCORPION-xisa [38]
3 years ago
12

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

A should have a frequency of 440 Hz and the note E should be at 659 Hz.
(a) What is the frequency difference between the third harmonic of the A and the second harmonic of the E?

(b) A tuner first tunes the A string very precisely by matching it to a 440 Hz tuning fork. She then strikes the A and E strings simultaneously and listens for beats between the harmonics. What beat frequency between higher harmonics indicates that the E string is properly tuned?

(c) The tuner starts with the tension in the E string a little low, then tightens it. What is the frequency of the E string when she hears four beats per second?
Physics
1 answer:
Sophie [7]3 years ago
8 0

(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

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(a) (i) Find the gradient of f. (ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rat
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Question:

Problem 14. Let f(x, y) = (x^2)y*(e^(x−1)) + 2xy^2 and F(x, y, z) = x^2 + 3yz + 4xy.

(a) (i) Find the gradient of f.

(ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rate is f decreasing?

(b) (i) Find the gradient of F.

(ii) Find the directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k.

Answer:

The answers to the question are

(a) (i)  the gradient of f =  ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) The direction in which f decreases most rapidly at the point (1, −1), ∇f(x, y) = -1·i -3·j is the y direction.

The rate is f decreasing is -3 .

(b) (i) The gradient of F is (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k is  ñ∙∇F =  4·x +⅟4 (8-3√3)y+ 9/4·z at (1, 1, −5)

4 +⅟4 (8-3√3)+ 9/4·(-5) = -6.549 .

Explanation:

f(x, y) = x²·y·eˣ⁻¹+2·x·y²

The gradient of f = grad f(x, y) = ∇f(x, y) = ∂f/∂x i+  ∂f/∂y j = = (∂x²·y·eˣ⁻¹+2·x·y²)/∂x i+  (∂x²·y·eˣ⁻¹+2·x·y²)/∂y j

= ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) at the point (1, -1) we have  

∇f(x, y) = -1·i -3·j  that is the direction in which f decreases most rapidly at the point (1, −1) is the y direction.  

The rate is f decreasing is -3

(b) F(x, y, z) = x² + 3·y·z + 4·x·y.

The gradient of F is given by grad F(x, y, z)  = ∇F(x, y, z) = = ∂f/∂x i+  ∂f/∂y j+∂f/∂z k = (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2·i + 3·j −√3·k

The magnitude of the vector 2·i +3·j -√3·k is √(2²+3²+(-√3)² ) = 4, the unit vector is therefore  

ñ = ⅟4(2·i +3·j -√3·k)  

The directional derivative is given by ñ∙∇F = ⅟4(2·i +3·j -√3·k)∙( (2·x+4·y)i + (3·z+4·x)j + 3·y·k)  

= ⅟4 (2((2·x+4·y))+3(3·z+4·x)- √3∙3·y) = 4·x +⅟4 (8-3√3)y+ 9/4·z at point (1, 1, −5) = -6.549

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Answer:

a)  Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a)  We have equation of motion s=ut+\frac{1}{2} at^2, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8m/s^2

 Substituting

    6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s

 Initial speed of the ball = 14.45 m/s

 b) We have equation of motion v^2=u^2+2as, where v is the final velocity

   s = 6 m, u = 14.45 m/s, a = -9.8m/s^2

    Substituting

        v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s

  So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion v^2=u^2+2as, where v is the final velocity

   u = 14.45 m/s, v =0 , a = -9.8m/s^2

   Substituting

     0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m

  Maximum height reached = 10.65 m

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The frequency of rotation of Mars is 0.0000113 Hertz.

<u>Given the following data:</u>

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To find the frequency of rotation in Hertz:

First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.

<u>Conversion:</u>

1 day = 24 hours

24 hours to minutes = 60 × 24 = 1440 minutes

1440 + 37 = 1477 \; minutes

1 minute = 60 seconds

1477 minute = X seconds

Cross-multiplying, we have:

X = 60 × 1477

X = 88620 seconds

Now, we can find the frequency of rotation of Mars by using the formula:

Frequency = \frac{1}{Period}\\\\Frequency = \frac{1}{88620}

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Therefore, the frequency of rotation of Mars is 0.0000113 Hertz.

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Answer:

They are made of one type of atom.

Explanation:

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