WAVE MC 1 題

Let n be the number of "loops" formed in the stationary wave when the string is vibrating at 400 Hz. Since one wavelength 入 contains 2 loops, or length of one "loop' equals to half a wavelength, hence in a string length of 1 m, we have,
n(入/2) = 1
i.e. 入 = 2/n

But v = f.入
where v is the speed of wave on the string, f is the wave frequency and 入 is the wavelength.
v = 400(2/n) ----------- (1)

Consider the frequency 450 Hz. Since there is no intermediate frequency between 400 Hz and 450 Hz, there are (n+1) loops when the string is vibrating at 450 Hz.
Thus, (n+1).(入'/2) = 1 where 入' is the new wavelength
入' = 2/(n+1)
Now, v = 450入' = 450[2/(n+1)] --------------------- (2)

Equating (1) and (2),
400(2/n) = 450[2/(n+1)]
8/n =9/(n+1)
9n = 8n + 8
i.e. n = 8

Substitute the value of n into (1)
v = 400 x 2/8 m/s = 100 m/s


,

2012-11-10 12:17:13 補充：
A displacement node is the point where the particle is always stationary (i.e. no displacement at all). Hence, a vibrating loop is formed between two sucessive displacement nodes.

2012-11-10 12:17:56 補充：
cont'd...
You may refer to the web-page below for displacement node in longitudinal wave:
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/standw.html