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Physics help Mass per unit length vs frequency?
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BackLumbarJack
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#1
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I am currently studying standing waves and harmonics, and am of the understanding that a high mass per unit length decreases the wave frequency on the string. Why would this happen? Does a high mass per unit length equate to a high density ( generally) and dont mechanical waves, such as those along a string, travel faster in denser media?
I am currently studying standing waves and harmonics, and am of the understanding that a high mass per unit length decreases the wave frequency on the string. Why would this happen? Does a high mass per unit length equate to a high density ( generally) and dont mechanical waves, such as those along a string, travel faster in denser media?
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Zangoose
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#2
(Original post by BackLumbarJack)
Hi
I am currently studying standing waves and harmonics, and am of the understanding that a high mass per unit length decreases the wave frequency on the string. Why would this happen? Does a high mass per unit length equate to a high density ( generally) and dont mechanical waves, such as those along a string, travel faster in denser media?
Hi
I am currently studying standing waves and harmonics, and am of the understanding that a high mass per unit length decreases the wave frequency on the string. Why would this happen? Does a high mass per unit length equate to a high density ( generally) and dont mechanical waves, such as those along a string, travel faster in denser media?
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BobBobson
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#3
(Original post by Zangoose)
Yeah, pretty much. You need to take the square root of the mass (divided by the mean of the wave frequency), then you take the exponential and you divide by the sum of the amplitude of each wave peak and then you end up with something that's proportionate to the square of the initial equation
Yeah, pretty much. You need to take the square root of the mass (divided by the mean of the wave frequency), then you take the exponential and you divide by the sum of the amplitude of each wave peak and then you end up with something that's proportionate to the square of the initial equation
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BackLumbarJack
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#4
(Original post by Zangoose)
Yeah, pretty much. You need to take the square root of the mass (divided by the mean of the wave frequency), then you take the exponential and you divide by the sum of the amplitude of each wave peak and then you end up with something that's proportionate to the square of the initial equation
Yeah, pretty much. You need to take the square root of the mass (divided by the mean of the wave frequency), then you take the exponential and you divide by the sum of the amplitude of each wave peak and then you end up with something that's proportionate to the square of the initial equation
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