Quick standing waves question
Paper:
http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/sam-gce-physics.pdf
I need some help on page 38, question 16b.
I've done the following working:
c=f*w so 1mx60Hz gives a speed of 60ms^-1.for the original setup.
1m=2pi radians
If 2 more nodes are added, then the cycle is 5pi radians so the scale factor is 5/2. Divide 1m by 5/2 to get 0.4m as being the new wavelength. speed/wavelength=frequency so 60/0.4=150Hz. Where have I gone wrong?
http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/sam-gce-physics.pdf
I need some help on page 38, question 16b.
I've done the following working:
c=f*w so 1mx60Hz gives a speed of 60ms^-1.for the original setup.
1m=2pi radians
If 2 more nodes are added, then the cycle is 5pi radians so the scale factor is 5/2. Divide 1m by 5/2 to get 0.4m as being the new wavelength. speed/wavelength=frequency so 60/0.4=150Hz. Where have I gone wrong?
Original post by bobbricks
Paper:
http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/sam-gce-physics.pdf
I need some help on page 38, question 16b.
I've done the following working:
c=f*w so 1mx60Hz gives a speed of 60ms^-1.for the original setup.
1m=2pi radians
If 2 more nodes are added, then the cycle is 5pi radians so the scale factor is 5/2. Divide 1m by 5/2 to get 0.4m as being the new wavelength. speed/wavelength=frequency so 60/0.4=150Hz. Where have I gone wrong?
http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/sam-gce-physics.pdf
I need some help on page 38, question 16b.
I've done the following working:
c=f*w so 1mx60Hz gives a speed of 60ms^-1.for the original setup.
1m=2pi radians
If 2 more nodes are added, then the cycle is 5pi radians so the scale factor is 5/2. Divide 1m by 5/2 to get 0.4m as being the new wavelength. speed/wavelength=frequency so 60/0.4=150Hz. Where have I gone wrong?
Not sure where you are getting that scale factor from. If 2 nodes are added, then we have 5 nodes instead of 3, so the scale factor would be 5/3.
Posted from TSR Mobile
Original post by Asklepios
Not sure where you are getting that scale factor from. If 2 nodes are added, then we have 5 nodes instead of 3, so the scale factor would be 5/3.
Posted from TSR Mobile
Posted from TSR Mobile
I thought there were 2 nodes and 3 antinodes?
Original post by bobbricks
I thought there were 2 nodes and 3 antinodes?
Well in this case you could argue there are 2, 3, or 4 depending on if you count the ones at the start/end or not. But, it is the sum of the distances between 3 nodes that makes up the 1.5m.
Original post by bobbricks
Paper:
http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/sam-gce-physics.pdf
I need some help on page 38, question 16b.
I've done the following working:
c=f*w so 1mx60Hz gives a speed of 60ms^-1.for the original setup.
1m=2pi radians
If 2 more nodes are added, then the cycle is 5pi radians so the scale factor is 5/2. Divide 1m by 5/2 to get 0.4m as being the new wavelength. speed/wavelength=frequency so 60/0.4=150Hz. Where have I gone wrong?
http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/sam-gce-physics.pdf
I need some help on page 38, question 16b.
I've done the following working:
c=f*w so 1mx60Hz gives a speed of 60ms^-1.for the original setup.
1m=2pi radians
If 2 more nodes are added, then the cycle is 5pi radians so the scale factor is 5/2. Divide 1m by 5/2 to get 0.4m as being the new wavelength. speed/wavelength=frequency so 60/0.4=150Hz. Where have I gone wrong?
In the 1st case you are correct with the wave like this
<><><>
This is 4 nodes and 3 antinodes.
And one wavelength is 1m as the string is 1.5m in length. (One wave marked in bold)
If you add "2 more nodes" you get
<><><><><>
which is 6 nodes and 5 antinodes.
As the wire is still 1.5m in length, one wavelength is now a smaller fraction of 1.5m and is marked in bold. Find this wavelength and plug it in v=fλ
(edited 10 years ago)
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