A sonometer wire is vibrating at frequency fo. If the tension in the wire is doubled while the length and the mass per unit length are kept constant, the new frequency of vibration is
- fo/2
- 2fo
- fo/√2
-
fo√2
The correct answer is: D
Explanation
For a vibrating sonometer wire, the frequency
F = l/2l √(T/m)
and if the length, l and mass per unit length, m, are constant, then Fo ∝ √T => FoK√T or Fo/√T = K
=> F1o/√F1 = F2o/√F2
∴ F1o/√T1 = F2o/√T2
| thus F2 = | F3x√(2T1 |
| __√T1 |
=
| = | Fox√2 x √T1 |
| √T1 |
∴F2 = Fo x √2 = Fox√2
There is an explanation video available .