Question

A block with mass m = 0.4 kg oscillates with amplitude A = 0.4 m at the end of a spring with force constant k = 15 N/m on a frictionless, horizontal surface. Rank the periods of the following oscillating systems from greatest to smallest. If any periods are equal, show their equality in your ranking. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.)
(a) The system is as described above.
(b) The amplitude is changed to 1.6 m.
(c) The mass is changed to 1.6 kg.
(d) The spring now has a force constant of 30 N/m.
(e) A small resistive force is added so the motion is underdamped.

Answer 1

Answer:

c > a= b >d > e

Explanation:

In a spring mass oscillatory system the angular velocity is given by

    w = √ k / m

With angular kinematics

    w = 2π f = 2π / T

Substituting

    2π / T = √ k / m

    T = 2π √ m / k

We fear the period equation let's see with change for different situations

a) m = 0.4 kg, k = 15 N / m

    T = 2π √A (0.4 / 15)

    T = 1.03 s

b) A = 1.5 m

The period does not change since it does not depend on the amplitude of the movement

c) The mass changes to m2 = 1.6 kg

    T2 = 2 π √ (1.6 / 15)

    T2 = 2.05 s

d) the spring constant changes to k = 30 N / m

     T3 = 2 π π (0.4 / 30)

     T3 = 0.726 s

e) Small resistive force is added

In this case the angular velocity decreases with each oscillation, in the way

     w = √ (k/m - (b/2m)²)

Since the period and angular velocity are inversely proportional, the period increases with each oscillation.