Answer:
The thicker wire will have a lesser frequency when producing standing waves
Explanation:
[tex]f=\frac{1}{2l} \sqrt{\frac{T}{m} }[/tex]
where,
f = frequency of the fundamental mode in producing standing waves
l = resonating length
T = tension of the wire
m = linear density of the wire
So in a thicker wire the mass per a unit length (m) is higher than that of a thinner wire.
So if we get two wires under same tension which all other factors except thickness of the wire is identical the thicker wire will have a lesser frequency when producing standing waves.
This deduction is made by,
f ∝ [tex]\frac{1}{\sqrt{m} }[/tex]
