A0.20 kg object, attached to a spring with spring constant k = 10 n/m, is moving on a horizontal frictionless surface in simple harmonic motion of amplitude of 0.080 m. what is its speed at the instant when its displacement is 0.040 m? (hint: use conservation of energy.)

 v = 0.566 m / s

Explanation:

Let's use energy conservation at two points the most extreme point of maximum elongation and the point of interest

Initial maximum elongation

    Em₀ = = ½ k x²

At this point x = A = 0.080 m

Final point of interest

    = K = ½ m v²

Energy is conserved

   Em₀ =

   ½ k A² = ½ m v²

   v² = k / m A²

   v = √(k/m) A

Let's calculate

   v = √ (10 / 0.20) 0.080

   v = 0.566 m / s

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).