Physics help
The frequency of the first harmonic on a standing wave is f. The length and tension are doubled, what affect does this have on the frequency?
I’ve spend the best part of an hour on this and it’s a one marker. Please help, with an explication.
I’ve spend the best part of an hour on this and it’s a one marker. Please help, with an explication.
Original post by -Mathematician-
The frequency of the first harmonic on a standing wave is f. The length and tension are doubled, what affect does this have on the frequency?
I’ve spend the best part of an hour on this and it’s a one marker. Please help, with an explication.
I’ve spend the best part of an hour on this and it’s a one marker. Please help, with an explication.
I am assuming that you are talking about the string that is fixed at both ends. The first harmonic is given by
where L is the length of the string, T is the tension and μ is the “mass density” or mass per unit length.
I believe that you can work from here to find the answer that is required by the question.
I got roo2/2 but this is wrong? The length gives 1/2 then the tension fives root2 surely?
Original post by -Mathematician-
I got roo2/2 but this is wrong? The length gives 1/2 then the tension fives root2 surely?
What is wrong? Is this a MCQ or a qualitative question?
Not sure what you are writing about.
It would be better that you post the full question.
(edited 6 years ago)
Original post by -Mathematician-
Please brush up your maths skill and learn how to simplify the radical.
Spoiler
Original post by Eimmanuel
Please brush up your maths skill and learn how to simplify the radical.
Spoiler
how come if you let mew=m/l doesn't that change the answer I know mew is constant but is it not allowed to make this substitution?
Original post by IsaiahD
how come if you let mew=m/l doesn't that change the answer I know mew is constant but is it not allowed to make this substitution?
Yes, by letting μ = M/L. It will change the answer from A to B. If you want to consider the mass does not change when the length of the string is doubled, I am good with it.
As far as I know, if we double the length without keeping the μ constant but the mass is constant, we can break the string “easily”.
I can be wrong about the answer.
Original post by Eimmanuel
Yes, by letting μ = M/L. It will change the answer from A to B. If you want to consider the mass does not change when the length of the string is doubled, I am good with it.
As far as I know, if we double the length without keeping the μ constant but the mass is constant, we can break the string “easily”.
I can be wrong about the answer.
As far as I know, if we double the length without keeping the μ constant but the mass is constant, we can break the string “easily”.
I can be wrong about the answer.
im not sure tbh, its just when you use density in equations you are allowed to have It as mass/volume and continue with calculations like that so im unsure here im pretty sure you write M/L
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