COMEDK2025PhysicsOscillationsActual
A wire of negligible mass having uniform area of cross section 'A' and young modulus ' Y ' is used to suspend a point mass ' m '. The point mass executes simple harmonic motion in a vertical plane with a period ' T ', then the length of the wire is :
Options
- AL= T Y^2 A 4 m^2
- BL= T^2 Y A 4 ^2 m
- CL= T^2 Y A 4 m^2
- DL= T Y^2 A 4 ^2 m
Correct answer
B. L= T^2 Y A 4 ^2 m
Step-by-step solution
The wire acts as a spring with a spring constant k given by the formula k = YA L , where Y is the Young modulus, A is the area of cross-section, and L is the length of the wire. The time period T of a mass m executing simple harmonic motion attached to a spring of constant k is given by T = 2 m k . Squaring both sides of the equation, we get T^2 = 4 ^2 m k . Substituting the expression for k into the equation for T^2 , we have T^2 = 4 ^2 m (YA/L) . Rearranging the equation to solve for L , we get T^2 = 4 ^2 mL YA .