The characteristics of the standing wave we can find the backlash for the frequency of the wave when the string is shortened is:
- The new frequency is f = 657 Hz
<h3>How is a standing wave produced?</h3>
A standing wave is produced when a traveling wave meets an obstacle and bounces, the sum of the two waves results in a wave that does not propagate in space.
In the event that the obstacle is a fixed point, there is a node at this point. The expression for the length of the standing wave.
L = fundamental frequency
L = second harmonic
L = third harmonic
L = general term.
Where L is the length of the chord, lan the wavelength and n an integer.
Wave speed is related to wavelength and frequency.
v = λ f.
Let's substitute.
v =
They indicate that initially the string has a length of L₀ = 327 mm= 0.327m and the frequency is f₀ = 440 Hz.
v n = 2L₀ f₀
v n = 2 0.327 440
v n = 287.76
They indicate that the tension on the string do not changes and the speed of the wave depends only on the tension and the density of the string, therefore it is constant, we assume that the harmonic does not change either, therefore the new length.
v n = 2 L f
Let's substitute.
287.76 = 2 L f
f =
Let's calculate.
f =
f = 656.99 Hz
In conclusion with the characteristics of the standing wave we can find the backlash for the frequency of the wave when the string is shortened is:
- The new frequency is: f = 657 Hz
Learn more about standing waves here: brainly.com/question/17031219
Answer:
Explanation:
Pressure is defined as the thrust acting per unit area.
the pressure by a person wearing loafers is less than the pressure by a woman wearing high heels.
It is because the area of loafers is much more than the area of heels.
Thus, the woman hurts a lot than man.
Answer:
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Explanation:
Answer:
The required mass of ice is 12.5 kg.
Explanation:
Mass of hot tea, = 1.8 kg
Initial temperature of hot tea = 80°C = 353 K
Initial temperature of ice = 0.00°C = 273 K
Final temperature of mixture = 10°C = 283 K
Heat of fusion of ice, L = 334 kJ/kg
Specific heat capacity of tea, = Specific heat capacity of water = 4190 J/kg/K
Heat lost tea = Heat gained by ice
Δ = L + Δ
Δ = (L + Δ)
1.8 x 4190 x (353 - 283) = (334 + (4190 x (283 - 2730))
1.8 x 4190 x 70 = (334 + (4190 x 10))
527940 = 42234
=
= 12.5004
= 12.5 kg
The required mass of ice is 12.5 kg.