Question

A block with mass m = 0.2 kg oscillates with amplitude A = 0.3 m at the end of a spring with force constant k = 12 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.
(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:

Explanation:

Given that,

A block of mass m = 0.2kg

Amplitude of oscillation A = 0.3m

Spring Constant k = 12N/m

We want to rank the period of oscillation in each case from the smallest to the largest.

Period of oscillation can be determine using

T = 2π√m/k

Where,

T is period in seconds

m is mass in kg

k is spring constant in N/m

So, the only things that affect the period is mass and the spring constant

A. Using the above information

Where m = 0.2kg and k = 12 N/m

Then, T = 2π√m/k

T = 2π√(0.2/12)

T = 2π × 0.129

Ta = 0.81 seconds

B. The amplitude is change to 1.6m, A = 1.6m

The period T for a pendulum is nearly independent of amplitude.

Since the period is independent of the amplitude, then, the period does not change

So, Tb = Ta = 0.81seconds

C. If the mass is change to 1.6kg

Now, m=1.6kg

Then, T = 2π√m/k

T = 2π√(1.6/12)

T = 2π × 0.365

Tc = 2.28 seconds

D. If the force constant is change to 30N/m

Now, k = 30

Then, T = 2π√m/k

T = 2π√(0.2/ 30)

T = 2π × 0.0816

Td = 0.513 seconds

E. The small resistive force does not affect the period

So the period remains unchanged

Ta = Te = 0.81 seconds

Ranking the periods

Tc > Ta = Tb = Te > Td