A Level physics- Waves question (1)
The length of PQ is 0.40 m. When the wire is vibrating, transverse waves are propagated along the wire at a speed of 64 m s–1. Explain why the wire is set into large amplitude vibration when the frequency of the a.c. supply is 80 Hz.
The Markscheme says:
wavelength (λ) of waves = 64/80 = 0.80 (m) (1)
length of wire is λ/2 causing fundamental vibration (1)
[or λ of waves required for fundamental (= 2 × 0.40) = 0.80 m (1)
natural frequency of wire = 80 (Hz) (1)]
wire resonates (at frequency of ac supply) [or a statement that
fundamental frequency (or a natural frequency) of the wire is
the same as applied
frequency] (1)
-For this question I understand that the main point is that I had to prove the applied frequency = the natural frequency and so therefore there is resonance and so the large amplitude
-BUT the problem I am having in understanding what the mark scheme has done is that when calculating the wavelength of the string it already assumed the applied frequency of 80Hz from the motor is the natural frequency of the wave no ?
The Markscheme says:
wavelength (λ) of waves = 64/80 = 0.80 (m) (1)
length of wire is λ/2 causing fundamental vibration (1)
[or λ of waves required for fundamental (= 2 × 0.40) = 0.80 m (1)
natural frequency of wire = 80 (Hz) (1)]
wire resonates (at frequency of ac supply) [or a statement that
fundamental frequency (or a natural frequency) of the wire is
the same as applied
frequency] (1)
-For this question I understand that the main point is that I had to prove the applied frequency = the natural frequency and so therefore there is resonance and so the large amplitude
-BUT the problem I am having in understanding what the mark scheme has done is that when calculating the wavelength of the string it already assumed the applied frequency of 80Hz from the motor is the natural frequency of the wave no ?
Not exactly.
The wavelength for this wave would be 0.8m and since the string is 0.4m. It has to look like this:
Thus it IS the fundamental frequency.
The wavelength for this wave would be 0.8m and since the string is 0.4m. It has to look like this:
Thus it IS the fundamental frequency.
Original post by mosaurlodon
Not exactly.
The wavelength for this wave would be 0.8m and since the string is 0.4m. It has to look like this:
Thus it IS the fundamental frequency.
The wavelength for this wave would be 0.8m and since the string is 0.4m. It has to look like this:
Thus it IS the fundamental frequency.
Thank you for taking the time to reply btw!
(edited 1 month ago)
Not really.
The wavelength was calculated using v=f*lambda
so you get 0.8m.
Which happens to be twice the string length.
That should be the logical sequence.
Thus it is the first harmonic.
The wavelength was calculated using v=f*lambda
so you get 0.8m.
Which happens to be twice the string length.
That should be the logical sequence.
Thus it is the first harmonic.
Original post by mosaurlodon
Not really.
The wavelength was calculated using v=f*lambda
so you get 0.8m.
Which happens to be twice the string length.
That should be the logical sequence.
Thus it is the first harmonic.
The wavelength was calculated using v=f*lambda
so you get 0.8m.
Which happens to be twice the string length.
That should be the logical sequence.
Thus it is the first harmonic.
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