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?
that is 1/4 of a period, going frm zero to max, right?
Yes it is.
It's actually 1/5 of a period?
To find the period of oscillation for a particle in Simple Harmonic Motion (SHM), we can use the formula:
T = 2π / ω
where:
T is the period of oscillation,
π is the mathematical constant (approximately 3.14),
and ω is the angular frequency.
In SHM, the time taken for a particle to travel from the equilibrium position to the maximum displacement (amplitude) and back again is half the period.
Given that it takes 0.20 seconds for the particle to travel from the equilibrium position to the maximum displacement, we can calculate the period.
Step 1: Divide the given time by 2 to get the time period for one-half of the oscillation.
T/2 = 0.20 s / 2 = 0.10 s
Step 2: Use the formula T = 2π / ω and solve for T.
0.10 s = 2π / ω
Step 3: Rearrange the equation to solve for ω.
ω = 2π / 0.10 s
Step 4: Calculate ω.
ω = 2π / 0.10 s ≈ 62.83 rad/s
Step 5: Substitute the value of ω back into the formula to find the period.
T = 2π / (62.83 rad/s)
Step 6: Simplify the equation to obtain the period.
T ≈ 0.10 s
Therefore, the period of oscillation is approximately 0.10 seconds.