F.5 Physics Homework

When the wire vibrates at fundamental frequency, the two fixed ends are always stationary (nodes), hence wavelength w = 2 x length of wire L
i.e. w = 2 x 5 m = 10 m
hence, speed of sound on the wire = 10 x 50 m/s = 500 m/s

Let x be the distance from the centre of wire where the bridge is placed.
Conisder the long segment:
length of wire segment = (5/2 + x) m = (2.5 + x) m
wavelength when vibrating at fundamental frequency w1 = 2(2.5+x) m
hence, fundamental frequency f1 = 500/[2(2.5+x)] Hz

Conisder the short segment:
length of wire segment = (5/2 - x) m = (2.5 - x) m
wavelength when vibrating at fundamental frequency w2 = 2(2.5-x) m
hence, fundamental frequency f2 = 500/[2(2.5-x)] Hz

Since it is given that 4 beats per second are heard,
i.e. f2 - f1 = 4
500/[2(2.5-x)] - 500/[2(2.5+x)] = 4
After simplification, we have
4x^2 + 500x - 25 = 0
solve for x and only accept the physically reasonable value of x gives x = 0.05 m (or 5 cm)

Therefore the bridge should be placed 5 cm from the centre of the wire.

MCQ question


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