A particle undergoes SHM has period of 0.4sec and amplitude of 12mm.Find the maximum velocity of the particle.
15.7 cm/sec
if y = a sin(kx) then
v = y' = ak cos(kx)
To find the maximum velocity of a particle undergoing Simple Harmonic Motion (SHM), we can use the formula:
v_max = Aω
Where:
v_max is the maximum velocity,
A is the amplitude of the motion, and
ω (omega) is the angular frequency.
First, let's find the angular frequency:
ω = 2π / T
Where:
T represents the period of the motion.
Given that the period of the motion is 0.4 seconds, we can find ω:
ω = 2π / 0.4
ω = 15.7 rad/s (rounded to two decimal places)
Now, we can calculate the maximum velocity:
v_max = A * ω
v_max = 12 mm * 15.7 rad/s
v_max ≈ 188.4 mm/s (rounded to one decimal place)
Therefore, the maximum velocity of the particle is approximately 188.4 mm/s.
To find the maximum velocity of a particle undergoing Simple Harmonic Motion (SHM), you can use the equation:
v_max = Aω
where v_max is the maximum velocity, A is the amplitude of the motion, and ω is the angular frequency.
To find ω, you can use the equation:
ω = 2π/T
where T is the period of the motion.
Given that the period of the SHM is 0.4 seconds (T = 0.4 sec) and the amplitude is 12 mm (A = 12 mm), let's calculate the maximum velocity step by step:
1. Convert the amplitude from millimeters to meters:
A = 12 mm = 0.012 m
2. Calculate the angular frequency (ω):
ω = 2π/T
ω = 2π/0.4 sec
ω ≈ 15.71 rad/s
3. Calculate the maximum velocity (v_max):
v_max = Aω
v_max = 0.012 m × 15.71 rad/s
v_max ≈ 0.18852 m/s
Therefore, the maximum velocity of the particle undergoing SHM is approximately 0.18852 m/s.