A particle that undergoes simple harmonic motion has a period of 0.4 s and an amplitude of 12 mm. The maximam velocity of the particle is?
4₹cm/s
To find the maximum velocity of a particle undergoing simple harmonic motion, you need to use the formula:
vmax = ω × A
Where:
vmax is the maximum velocity
ω (omega) is the angular frequency of the motion
A is the amplitude of the motion
First, we can find the angular frequency (ω) using the formula:
ω = 2π / T
Where T is the period.
Given:
Period (T) = 0.4 s
Amplitude (A) = 12 mm = 0.012 m
Using the formula for angular frequency:
ω = 2π / T = 2π / 0.4 = 15.71 rad/s
Now, we can calculate the maximum velocity:
vmax = ω × A = 15.71 rad/s × 0.012 m = 0.18852 m/s
Therefore, the maximum velocity of the particle is approximately 0.18852 m/s.
To find the maximum velocity of a particle undergoing simple harmonic motion, you can use the formula:
ω = 2π / T
where ω represents the angular frequency and T is the period of the motion.
First, let's calculate the angular frequency using the given period:
ω = 2π / T
= 2π / 0.4 s
= 5π rad/s
Next, we can find the maximum velocity using the formula:
v_max = A * ω
where v_max represents the maximum velocity and A is the amplitude of the motion.
Plugging in the given values:
v_max = 12 mm * 5π rad/s
≈ 60π mm/s
To get the maximum velocity in meters per second (m/s), we need to convert mm/s to m/s by dividing by 1000:
v_max ≈ (60π mm/s) / 1000
≈ 0.1885 m/s (rounded to four decimal places)
Therefore, the maximum velocity of the particle is approximately 0.1885 m/s.
y = .012 sin (2pi t/.4)
v = dy/dt = (.012)(2 pi/.4)cos (2 pi t/.4)
max v = .012 * 2 * pi / .4