If it takes a particle in SHM 0.20 s to travel from the equilibrium position to the maximum displacement (amplitude), what is the period of oscillation?
Four times as long, 0.80 s.
The period of oscillation in Simple Harmonic Motion (SHM) is the time it takes for a particle to complete one full cycle of motion. In this case, the time it takes for the particle to travel from the equilibrium position to the maximum displacement and back to the equilibrium position is given as 0.20 seconds.
To determine the period, we need to consider that the time for one full cycle includes both the motion from the equilibrium position to the maximum displacement and the motion from the maximum displacement back to the equilibrium position.
Therefore, the period (T) of oscillation can be calculated by multiplying the given time by 2:
T = 2 * 0.20 seconds
T = 0.40 seconds
So, the period of oscillation is 0.40 seconds.
To find the period of oscillation for a particle in Simple Harmonic Motion (SHM), we can use the relation between period and frequency.
The period (T) is the time taken for one complete oscillation, while the frequency (f) is the number of oscillations per unit time. The two are related by the equation:
T = 1 / f
In this case, we are given the time taken for the particle to travel from the equilibrium position to the maximum displacement, which is equal to half of the period (T/2). Let's call this time as t.
t = 0.20 s
Since t is half the period, we can write:
t = T / 2
Rearranging the equation, we get:
T = 2t
Plugging in the given value for t, we have:
T = 2 * 0.20 s
Calculating the value, we find:
T = 0.40 s
Therefore, the period of oscillation for the particle in SHM is 0.40 seconds.