# A level physics waves question

And d for this one
(edited 1 month ago)
Using the first harmonic equation, we know that frequency is directly proportional to root tension. This means 300 equals k root T, so we can say 300 over root T is equal to the new frequency over root 2T ( tension is doubling). Rearranging, new f equals 300 times root 2 (we can ignore the T's and just pay attention to their coefficients). This gives 424, approximately 420.
Original post by orthopedic-rabbi
Using the first harmonic equation, we know that frequency is directly proportional to root tension. This means 300 equals k root T, so we can say 300 over root T is equal to the new frequency over root 2T ( tension is doubling). Rearranging, new f equals 300 times root 2 (we can ignore the T's and just pay attention to their coefficients). This gives 424, approximately 420.

Sorry I don’t get why we do 300/root T=f/root 2T
(edited 1 month ago)
For the second one, if you touch the fubdemetal frequency one third of the way it becomes the third harmonic. This means it is one third of the initial wavelength, lambda over 3. Wavelength and frequency are inversely proportional, hence lambda over 3 is 3f.
Original post by Gcsestudent56
Sorry I don’t get why we do 300/root T=f/root 2T

Ok so we know that f is proportional to root T. We can then say f equals k root T, k being a constant. In the second scenario tension is being doubled, so this means the new f equals k root 2T. We can rearrange both for k and equate them, because k is going to be the same value in both equations. Hence we can say initial f over root T equals new f over root 2T
Original post by orthopedic-rabbi
For the second one, if you touch the fubdemetal frequency one third of the way it becomes the third harmonic. This means it is one third of the initial wavelength, lambda over 3. Wavelength and frequency are inversely proportional, hence lambda over 3 is 3f.

How does touching it lightly 1/3 of the way make it a third harmonic?
Original post by orthopedic-rabbi
Ok so we know that f is proportional to root T. We can then say f equals k root T, k being a constant. In the second scenario tension is being doubled, so this means the new f equals k root 2T. We can rearrange both for k and equate them, because k is going to be the same value in both equations. Hence we can say initial f over root T equals new f over root 2T

Ahh ok makes sense
Touching the string 1/3 of the way makes a node at that point. If you draw it out you can see that if there is a node 1/3 of the way it is the third harmonic and hence has triple the initial frequency. These questions are from pmt right? If you onto the worked solutions you can see it more visually in their diagram.
Original post by orthopedic-rabbi
Touching the string 1/3 of the way makes a node at that point. If you draw it out you can see that if there is a node 1/3 of the way it is the third harmonic and hence has triple the initial frequency. These questions are from pmt right? If you onto the worked solutions you can see it more visually in their diagram.

I think only full papers have worked solutions. I’m doing the topic questions
Original post by Gcsestudent56
I think only full papers have worked solutions. I’m doing the topic questions

The topic questions do have worked solutions I think, they are called MCQ MA
Original post by orthopedic-rabbi
The topic questions do have worked solutions I think, they are called MCQ MA
I was on the QP, I found it on the MC paper. All it shows is a diagram of a third harmonic. I do get why if there’s 2 nodes, it is a third harmonic and therefore x3 the fundamental frequency. What I can’t really visualise is why the stationary wave is being formed even after the string has been pressed down, wouldn’t pressing it down mean that 2 new fixed ends are formed which just makes the length of string at which the waves are travelling smaller? I might be waffling idk
Original post by Gcsestudent56
I was on the QP, I found it on the MC paper. All it shows is a diagram of a third harmonic. I do get why if there’s 2 nodes, it is a third harmonic and therefore x3 the fundamental frequency. What I can’t really visualise is why the stationary wave is being formed even after the string has been pressed down, wouldn’t pressing it down mean that 2 new fixed ends are formed which just makes the length of string at which the waves are travelling smaller? I might be waffling idk

The fixed ends are still the same as they were initially, but when you pinch the sting we assume it creates a node there. However the rest of the 2/3 of the string is still in tension hence a wave still oscillates meaning it is a third harmonic not just a smaller first harmonic. Does that make sense? For it to create a fixed end the rest of the string cannot be in tension.
Original post by orthopedic-rabbi
The fixed ends are still the same as they were initially, but when you pinch the sting we assume it creates a node there. However the rest of the 2/3 of the string is still in tension hence a wave still oscillates meaning it is a third harmonic not just a smaller first harmonic. Does that make sense? For it to create a fixed end the rest of the string cannot be in tension.

Yh I’m kinda getting it. Thanks for your help