Now Etienne and Hakim are ready to try a homework problem. A 114 g block connected to a light spring with a force constant of k = 4 N/m is free to oscillate on a horizontal, frictionless surface. The block is displaced 6 cm from equilibrium and released from rest. A. Find the period of its motion. (Recall that the period, T, and frequency, f, are inverses of each other.)
Simple harmonic motion
Simple harmonic motion is a type of periodic motion in which an object undergoes oscillatory motion. The restoring force exerted by the object exhibiting SHM is proportional to the displacement from the equilibrium position. The force is directed towards the mean position. We see many examples of SHM around us, common ones are the motion of a pendulum, spring and vibration of strings in musical instruments, and so on.
Simple Pendulum
A simple pendulum comprises a heavy mass (called bob) attached to one end of the weightless and flexible string.
Oscillation
In Physics, oscillation means a repetitive motion that happens in a variation with respect to time. There is usually a central value, where the object would be at rest. Additionally, there are two or more positions between which the repetitive motion takes place. In mathematics, oscillations can also be described as vibrations. The most common examples of oscillation that is seen in daily lives include the alternating current (AC) or the motion of a moving pendulum.
Now Etienne and Hakim are ready to try a homework problem.
A 114 g block connected to a light spring with a force constant of k = 4 N/m is free to oscillate on a horizontal, frictionless surface. The block is displaced 6 cm from equilibrium and released from rest.
A. Find the period of its motion. (Recall that the period, T, and frequency, f, are inverses of each other.)
B. Determine the maximum acceleration of the block.
Given
Mass of block = 114g = 114×10-3kg
Force constant k = 4 N/m
Distance = 6cm = 0.06 m
Let w = Omega
w = ✓k/m =( 4/114×10-3)1/2 = (4000/114)1/2 = (35.08)1/2 = 5.92 rad/sec
T= 2π /w = (2×3.14)/5.92 = 6.28/5.92 = 1.06s
Vmax= wA=5.92×0.06 = 0.355 = 0.36m/s
amax = w2A = (5.92)2 × 0.06
= 35.0464 × 0.06 = 2.102 m/s2
Trending now
This is a popular solution!
Step by step
Solved in 2 steps
