Question

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

Answer 1

answer : 0.39 J

explanation : 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

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]

Answer 2

In simple harmonic motion, there are two factors particularly there that are the kinetic energy and potential energy.

• It depends upon the displacement, if it is maximum then the potential energy is maximum and kinetic energy is zero.
• At the equilibrium potential energy is zero and kinetic energy is maximum.
• Therefor as linear oscillator mechanical energy.
• E=kx²/2
• As x=11, k= 65N/m
• Put in that E=65×11²/2
• The mechanical energy be 1966.25N/m