Question

the elastic potential energy of the spring is equal to the kinetic energy of the bob when the distance of the bob executing shm from the equilibrium position in 0.24m calculate the amplitude of the bob ​

Answer 1

• The elastic potential energy of a spring is given by the equation:

                                P.E. = 1/2*m*ω²*x²

• Where m  = mass of the bob
• ω =  Angular frequency
• x = displacement from equilibrium position
• The kinetic energy of a spring is given by :

                       K.E. = 1/2*m*ω²(A²-x²)

• Here A = Amplitude of the bob
• the elastic potential energy of the spring is equal to the kinetic energy of the bob when the distance of the bob executing SHM from the equilibrium position in 0.24 m
• Hence P.E. = K.E. when x = 0.24 m
•                        1/2*m*ω²*x² = 1/2*m*ω²*(A²-x²)
• or,                    x² = A² - x²
• or,                    A=√2*x²
• putting x= 0.24 m, we get A = 0.24√2 m ≈ 0.339 m