✔ 最佳答案
(1a) Since the loop width are 18 cm and 16 cm in consecutive modes, the length of the string must be a multiple of both 18 and 16.
Looking into the smallest possible multiplier of 18 and 16, we have:
The minimum possible length of the string = 144 cm (1.44 m)
(1b) Converting the linear mass density back to the SI unit: 0.004 kg/m:
Also applying the equation for travelling velocity of transverse wave on a stretched string:
v = √(T/μ) where T = tension and μ = linear mass density
v = √(10/0.004) = 50 m/s
So in the fundamental mode, L = λ/2, i.e. λ = 2L = 2.88 m
Frequency f = v / λ = 17.36 Hz
(2) First of all, we have to obtain the travelling wave velocity on the string first:
v = √(T/μ) = √(12/0.0026) = 67.9 m/s
Now, suppose that the nth harmonic and (n+1)th harmonic frequencies are 480 Hz and 600 Hz respectively, we have:
Wavelength at nth harmonic λn = 2L/(n+1)
Wavelength at (n + 1)th harmonic λn+1 = 2L/(n+2)
Then we have:
v/480 = 2L/(n+1) ... (1)
v/600 = 2L/(n+2) ... (2)
(1) ÷ (2) yields:
(n+2)/(n+1) = 600/480
1 + 1/(n+1) = 5/4
1/(n+1) = 1/4
n = 3
So 480 Hz is the third harmonic frequency which is 4 times the fundamental frequency.
So for part (A), the fundamental mode frequency is 120 Hz.
Then for part (B), we first find out the wavelength of the fundamental mode which is given by:
λ = v/f = 67.9/120 = 0.566 m
Then from the relation L = λ/2, the length of the string is 0.284 m.
