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)
Answers
Answer:
a= 0.39J b=0.30J
Explanation:
mechanical energy of spring-mass system is constant.
mechanical energy = kinetic energy + potential energy = constant
the block's position is x = 11cm
k = 65N/m
so, initial p. e = 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
initial kinetic energy = 1/2 mv² =
mechanical energy = kinetic energy + potential energy
= 0 + 0.39J
= 0.39J
(b) , x = A/2 , where A is amplitude of oscillations.
potential energy = 1/2 K(A/2)² = 1/8 KA²
when kinetic energy is zero, potential will be maximum and we know, potential will be maximum at height point which is amplitude.
getting potential energy in question (a) is highest potential energy.
U = 1/2 KA² = 0.39 J
we get, potential energy = 1/4 U = 0.39/4 = 0.0975 ≈ 0.98 J
now, kinetic energy = mechanical energy - potential energy
= 0.39 J - 0.0975 J = 0.2925 J ≈ 0.30J