MHT CET202015 Oct 2020Evening ShiftPhysicsOscillationsActual
A particle of mass ' m ' is executing simple harmonic motion about its mean position. If ' A^ is the amplitude and ' T ' is the period of S.H.M., then the total energy of the particle is
Options
- A4 ² ~mA ² ~T ²
- B8 ² ~mA ² ~T ²
- C2 ² ~mA ² ~T ²
- D² ~mA ² ~T ²
Correct answer
C. 2 ² ~mA ² ~T ²
Step-by-step solution
(x=A t, ,= d x d t = A t ) (T E=P E+K E, P₀ E= 1 2 k x², K E = 1 2 m v² ) T. (E= 1 2 k A² ² t+ 1 2 m ² A² ² t , = R m , k mv² ) T. (E= 1 2 m ² A² ² t+ 1 m ² A² c c² t 2 m ² A² 2 = 2 m ² A² T² )