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CHECK YOURSELF ANSWERS
1All points have the same frequency, C.
2The point in the middle since the point there moves the largest distance in one period.
3The speed doubles and the frequency stays the same so the wavelength doubles. The string will now have just one loop, the first harmonic.
4The wavelength is given by Î»=4Ln where n is odd and so the frequency is f=cÎ»=c4Ln.
So, 126=c4LnÂ andÂ 210=c4L(n+2).
This gives 210126=c4L(n+2)c4Ln=n+2n.
Thus, 53=n+2nâ5n=3n+6 and so n = 3.
This means 126=f1Ã3âf1=42Â Hz.
Alternatively, you can argue as follows:
The frequencies are given by f=c4Ln with n odd.
So, two consecutive harmonics differ in frequency by the constant amount C2Â L=210â126=84Â Hz.
Hence the first harmonic has frequency 126 â 84 = 42 Hz.
Yet another way to solve this is that the difference in frequency of two consecutive harmonics in the pipe is double the first harmonic frequency and so the answer is half of 84 Hz or 42 Hz.
And, finally, another way is to find the greatest common factor of 210 = {210, 70, 42, 30,...}, and 126 = {126, 42, 25.2, ...} i.e. 42 Hz.
5The wavelength is 4L where L is the length of the string. Thus, 24 = 5.0 Ã 4L â L = 1.2 m. (This is a simplistic answer; this is worth investigating as an IA perhaps.)
6The lower curve.
7From the discussion in the text, the answer is C.
