Question

 a) What is the mechanical energy E of the linear oscillator of Problem 1 above. (Initially, the block’s position is x = 11 cm and its speed is v = 0. Spring constant k is 65 N/m.)
b) What are the potential energy and kinetic energy of the oscillator when the block is at x = A/2? (A represents the amplitude of oscillations)​

Answer 1

(a) here you should remind that mechanical energy of spring-mass system always remains constant.

i.e., mechanical energy = kinetic energy + potential energy = constant

initially, the block's position is x = 11cm

given, k = 65N/m

so, initial potential energy = 1/2 kx²

= 1/2 × 65N/m × (11 × 10^-2 m)²

= 0.39325 J ≈ 0.39 J .....(1)

initial speed of block, v = 0

so, initial kinetic energy = 1/2 mv² = 0

hence, mechanical energy = kinetic energy + potential energy

= 0 + 0.39J

= 0.39J [Ans]

(b) as it is given that, x = A/2 , where A is amplitude of oscillations.

so, potential energy = 1/2 K(A/2)² = 1/8 KA² ......(2)

you know, when kinetic energy is zero, potential will be maximum and we know, potential will be maximum at height point.i.e., amplitude.

so, getting potential energy in question (a) is highest potential energy.

so, U = 1/2 KA² = 0.39 J

from equation (2),

we get, potential energy = 1/4 U = 0.39/4 = 0.0975 ≈ 0.98 J [ans]

now, kinetic energy = mechanical energy - potential energy

= 0.39 J - 0.0975 J = 0.2925 J ≈ 0.30J [Ans]