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numerical problems [4 marks]
30. 2076 set b q. no. 111 a wire whose mass per unit length
10^-3 kg/m is stretched by a load of 4 kg over the two bridges
of a sonometer wire 1 m apart. it is struck at its middle point,
what would be the wavelength and frequency of its
fundamental vibration?
ans: 100 hz
60
[4]

λ = 2 m

f = 100 Hz

Explanation:

When struck in the middle an anti-node is formed at the center, So you can derive,

f = frequency of the fundamental mode in producing standing waves

l = resonating length

T = tension of the wire

m = linear density of the wire

(check the attachment)

By substituting,

f = 100 Hz

Wavelength is twice the vibrating  length, its in the section standing waves and using that only the equation is derived.

Imagine what happens when two identical waves in opposite direction superimpose. There will be 3 nodes and two anti-nodes where the distance between two nodes is half the wavelength.

The same case happen here, the transverse sound wave traveling along the wire get reflected by a bridge and bounce back on itself where superposition takes place. So two nodes are at the bridges hence the twice of the distance between bridges is the wavelength of the sound wave.