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Original post by Kallisto
I have a question in terms of frequency: is a harmonic always higher than fundamental oscillation? that makes sense in my view, because harmonics have always more wavelengths than fundamental oscillations. So harmonics must have a higher frequency. The more the higher. Am I right?
Think of a standing wave on a string, confined at each end. The amplitude must be zero at each end and so the fundamental frequency is the lowest frequency you can have (with wavelength twice the length of the string). A Harmonic is then some multiple of the fundamental frequency.
Original post by F1 fanatic
Think of a standing wave on a string, confined at each end. The amplitude must be zero at each end and so the fundamental frequency is the lowest frequency you can have (with wavelength twice the length of the string). A Harmonic is then some multiple of the fundamental frequency.
Thanks ofr answer. you are helpful. And now I want just to know if there is an alternative deduction to the damped oscillation. In my considerations, I have used a differential calculus and the integral calculus to get the formula for spring pendulum. But its too complicated in my view. Moreover the evidence for Euler's number is too implausible. What is the best way to get the formula for damped oscillation in a spring pendulum?
Original post by Kallisto
Thanks ofr answer. you are helpful. And now I want just to know if there is an alternative deduction to the damped oscillation. In my considerations, I have used a differential calculus and the integral calculus to get the formula for spring pendulum. But its too complicated in my view. Moreover the evidence for Euler's number is too implausible. What is the best way to get the formula for damped oscillation in a spring pendulum?
As far as I can remember you can't do it mathematically without some reference to calculus. It's a second order differential equation, which is fairly standard to solve, usually be assuming a form for the solution, inserting it into the equation and then determining the constants.
Original post by F1 fanatic
(...) and then determining the constants.
What did you mean? damping constant? Euler's number? As far as I know the damping constant is important to distinguish between three oscillators: overdamped case, critical damping and underdamped case.
(edited 11 years ago)
Original post by Kallisto
What did you mean? damping constant? Euler's number? As far as I know the damping constant is important to distinguish between three oscillators: overdamped case, critical damping and underdamped case.
See if you can follow this http://www.haverford.edu/physics-astro/dcross/academics/misc/dho.pdf
Original post by F1 fanatic
See if you can follow this http://www.haverford.edu/physics-astro/dcross/academics/misc/dho.pdf
Thanks for link. As like I have to use differential and integral calculus to get it.
Original post by Kallisto
Thanks for link. As like I have to use differential and integral calculus to get it.
Differentiation yes, integration no. Calculus is by far the easiest way of deriving it, as far as I know. You best get used to it, should you do a physics degree you will spend your life solving differential equations.
I thought about the spring pendulum. Displacement and amplitude s0 to be exactly. In my consideration the spring pendulum forces both gravitation and resilience. So this is my equation:
Fg = Fr
m*g = D*s => s = m*g / D.
That would be the displacement of the spring pendulum.
To calculate the amplitude s0, I thought about the potential relationships. In my consideration the total energy is the same to all kinds of moments in terms of the displacement of the spring pendulum:
E = m*v².
In my view the velocity v is the maximum one, vm. The maximum velocity can be calculate by the maximum displacement sm which is equal to the displacement s at the beginning and the circuit frequency, omega:
vm = s*omega; omega = 2*pi/T
So, the calculation of total energy is: E = m*(s*2*pi/T)². And when I get the total energy, I'm able to calculate the amplitude s0 which kinetic energy is 0 (due to the moment t = 0). In my consideration the total energy consists of elongation. That's why my equation is:
E = 1/2*D*s0² => s0 = square root of 2*E/D
But if I'm consider the amplitude s0 is the maximum one, and the maximum is the displacement at the beginning, my calculation for displacement must be equivalent to the calculation for maximum amplitude:
s = s0 = m*g/D = square root of 2*E/D.
Am I right with all my considerations?
Fg = Fr
m*g = D*s => s = m*g / D.
That would be the displacement of the spring pendulum.
To calculate the amplitude s0, I thought about the potential relationships. In my consideration the total energy is the same to all kinds of moments in terms of the displacement of the spring pendulum:
E = m*v².
In my view the velocity v is the maximum one, vm. The maximum velocity can be calculate by the maximum displacement sm which is equal to the displacement s at the beginning and the circuit frequency, omega:
vm = s*omega; omega = 2*pi/T
So, the calculation of total energy is: E = m*(s*2*pi/T)². And when I get the total energy, I'm able to calculate the amplitude s0 which kinetic energy is 0 (due to the moment t = 0). In my consideration the total energy consists of elongation. That's why my equation is:
E = 1/2*D*s0² => s0 = square root of 2*E/D
But if I'm consider the amplitude s0 is the maximum one, and the maximum is the displacement at the beginning, my calculation for displacement must be equivalent to the calculation for maximum amplitude:
s = s0 = m*g/D = square root of 2*E/D.
Am I right with all my considerations?
"The scientists of today think deeply instead of clearly. One must be sane to think clearly, but one can think deeply and be quite insane."
(edited 11 years ago)
Original post by boromir9111
"The scientists of today think deeply instead of clearly. One must be sane to think clearly, but one can think deeply and be quite insane."
"The scientists of today think deeply instead of clearly. One must be sane to think clearly, but one can think deeply and be quite insane."
Thanks Boz.
Original post by dknt
thanks boz.
brudahhhhhhh
Hey, can I join this society please?
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This was posted from The Student Room's iPhone/iPad App
Original post by Naadirah21
Hey, can I join this society please?
This was posted from The Student Room's iPhone/iPad App
This was posted from The Student Room's iPhone/iPad App
Click HERE and join
I have a question in terms of displacement. Is the maximum displacement the way which deflect the distance of a spring pendulum when the weight will be add to the one? Then this is my formula for maximum displacement:
sm = s = m*g /D.
Here are some explanations, if you are confuse:
sm: maximum displacement
s: displacement
m: mass (of the weight)
g: gravity (9,8 N)
D: spring rate
sm = s = m*g /D.
Here are some explanations, if you are confuse:
sm: maximum displacement
s: displacement
m: mass (of the weight)
g: gravity (9,8 N)
D: spring rate
Original post by boromir9111
Click HERE and join
I can't seem to find the option to join... Whereabouts is it on the page?
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Original post by Naadirah21
I can't seem to find the option to join... Whereabouts is it on the page?
This was posted from The Student Room's iPhone/iPad App
This was posted from The Student Room's iPhone/iPad App
Should be top right in an orange box.
Original post by Naadirah21
I can't seem to find the option to join... Whereabouts is it on the page?
This was posted from The Student Room's iPhone/iPad App
This was posted from The Student Room's iPhone/iPad App
What the above poster said
I'm already a member of TSR, I have an account etc... The orange box says 'sign up for free'
This was posted from The Student Room's iPhone/iPad App
This was posted from The Student Room's iPhone/iPad App
Original post by Naadirah21
I'm already a member of TSR, I have an account etc... The orange box says 'sign up for free'
This was posted from The Student Room's iPhone/iPad App
This was posted from The Student Room's iPhone/iPad App
If you're seeing that then you're not logged in. Once you have logged in, it's not that far up the page in the link provided by boromir (see screenshot), to the right of where it says 'Physics Society'
Original post by F1 fanatic
If you're seeing that then you're not logged in. Once you have logged in, it's not that far up the page in the link provided by boromir (see screenshot), to the right of where it says 'Physics Society'
Thank you
I've joined now.This was posted from The Student Room's iPhone/iPad App
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