NTA Abhyas JEE Main2020PhysicsUnits and DimensionsPractice
The frequency of vibration f of a mass m suspended from a spring of spring constant k is given by a relation of the type f = c m x k y , where c is a dimensionless constant. The values of x and y are
Options
- Ax = 1 2 , y = 1 2
- Bx = - 1 2 , y = - 1 2
- Cx = 1 2 , y = - 1 2
- Dx = - 1 2 , y = 1 2
Correct answer
D. x = - 1 2 , y = 1 2
Step-by-step solution
f = c   m x   k y ; Spring constant k =   f o r c e / l e n g t h . M 0   L 0   T - 1 = M x MT - 2 y = M x + y   T - 2 y ⇒ x + y = 0 ,   - 2 y =   - 1   or y = 1 2 Therefore, x =   - 1 2