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Hi, I've been stuck on this question & how the answer is correct. Here it is:
Question:
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.
Answer:
wavelength (λ) of waves = = 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)
How's this correct?
Question:
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.
Answer:
wavelength (λ) of waves = = 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)
How's this correct?
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#3
Hey there this is a note I came across 1 day ago and it is helping me clear my doubts in waves lessons but prob is it's a new website and still being developing but fast have a check
https://notesnsolutions.com/cie-a-level-waves-as/
https://notesnsolutions.com/cie-a-level-waves-as/
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#4
the ms is right. if a stationary wave vibrates with the fundamental frequency, you only see half the vibrating section. so if the length of the string is 0.4m the wavelength is 0.8m. as wavespeed = frequency x wavelength, 64/0.8 = 80Hz. resonance occurs when the natural frequency of the system = frequency of the forced vibrations. so as 80Hz is the natural frequency and the resonant frequency, large oscillations occur at this frequency. if you don't get this explanation you probably need to look at some of the content before you start answering questions, just some advice. 
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(Original post by Harry_Spencer)
the ms is right. if a stationary wave vibrates with the fundamental frequency, you only see half the vibrating section. so if the length of the string is 0.4m the wavelength is 0.8m. as wavespeed = frequency x wavelength, 64/0.8 = 80Hz. resonance occurs when the natural frequency of the system = frequency of the forced vibrations.s th so as 80Hz ie natural frequency and the resonant frequency, large oscillations occur at this frequency. if you don't get this explanation you probably need to look at some of the content before you start answering questions, just some advice.
the ms is right. if a stationary wave vibrates with the fundamental frequency, you only see half the vibrating section. so if the length of the string is 0.4m the wavelength is 0.8m. as wavespeed = frequency x wavelength, 64/0.8 = 80Hz. resonance occurs when the natural frequency of the system = frequency of the forced vibrations.s th so as 80Hz ie natural frequency and the resonant frequency, large oscillations occur at this frequency. if you don't get this explanation you probably need to look at some of the content before you start answering questions, just some advice.
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