Answer:
The thicker wire will have a lesser frequency when producing standing waves
Step-by-step explanation:
![f=(1)/(2l) \sqrt{(T)/(m) }](https://img.qammunity.org/2020/formulas/physics/middle-school/nty4q1fb7t6qbxs7f5c5hqp6g2iif7bbp5.png)
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 ∝
![(1)/(√(m) )](https://img.qammunity.org/2020/formulas/physics/middle-school/e8cuyj1lb7qxkvdvrmuo973ka69ps3igh6.png)