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Elis [28]
3 years ago
11

If a particle undergoes shm with amplitude 0.21 m, what is the total distance it travels in one period?

Physics
1 answer:
Ymorist [56]3 years ago
7 0
If a particle undergoes simple harmonic motion with an amplitude of 0.21 meters, this means that the maximum displacement of the particle from its resting position is 0.21. For one period, it traveled from its starting position which is twice the amplitude and then back to its original position which is another distance that is twice the amplitude as well. Therefore, the total distance it traveled is 2*amplitude + 2*amplitude = 2*0.21 + 2*0.21 = 0.42 + 0.42 = 0.84 meters.
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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
2 years ago
The demand for your hand-made skateboards, in weekly sales, is q = −3p + 700 if the selling price is $p. You are prepared to sup
Y_Kistochka [10]

Answer:

you should sell your skateboards at $240

Explanation:

The price p to sell your skateboards for so that there is neither a shortage nor a surplus is the price that makes equal the quantity of sales and the quantity of supply, so p is equal to:

q (sales) = q (supply)

-3p + 700 = 2p - 500

700 + 500 = 2p + 3p

1200 = 5p

1200/5 = p

$240 = p

3 0
3 years ago
Two objects attract each other with a gravitational force of magnitude 1.02 10-8 N when separated by 19.7 cm. If the total mass
Yakvenalex [24]

Answer:

mass 1 = 1.60kg

mass 2 = 3.54kg

4 0
3 years ago
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I WILL MARK YOU THE BRAINLIEST NO LINKS
aliya0001 [1]

Answer:

it is 30 newtons to the left

Explanation:

6 0
3 years ago
A horizontal force of 350N is exerted on a 2.5 kg ball as it rotates uniformly in a horizontal circle of radius of 0.90m. Calcul
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F=mv^2/R
----> V^2=FR/m=(350x0.9)/2.5=126
----- V=11.22 m/s
5 0
3 years ago
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