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A particle moving along the x-axis in simple harmonic motion starts from its equilibrium position, the origin, at t=0 and moves to the right. The amplitude of its motion is 2.00cm and the frequency is 1,50 HZ

a) show that the position of the particle is given by
x=(2.00cm) sin (3.00 πt)
Determine
b) the maximum speed and the earliest time (t>0) at which the particle has this speed,
c) the maximum acceleration and the earliest time (t>0 at which the particle has this acceleration
d) the total distance traveled between t=0 and t=1.00S.

User Pw
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1 Answer

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The particle's position is given by x = (2.00 cm) sin(3.00 πt).

Maximum speed = 18.84 cm/s at t = 0, 0.667 s, 1.333 s, etc.

Maximum acceleration = 565.5 cm/s² at t = 0.167 s, 0.833 s, 1.500 s, etc.

Total distance traveled between t = 0 and t = 1.00 s is ≈ 1200.06 cm.

How can you show what the position of the particle is given by?

In simple harmonic motion with amplitude A and frequency f, the position of the particle at any time t is given by:

x = A sin(2πft)

Here, A = 2.00 cm and f = 150 Hz = 150 π rad/s (since f = ω/2π). Therefore, the equation for the particle's position is:

x = (2.00 cm) sin(3.00 πt)

b) The maximum speed in simple harmonic motion occurs at the equilibrium position (x = 0). The velocity can be obtained by differentiating the position equation:

v = dx/dt = (6.00 π rad/s) cos(3.00 πt)

Maximum speed occurs when cos(3.00 πt) = 1, which is at t = 0, 0.667 s, 1.333 s, etc. (multiples of 1/3 cycle). Therefore, the maximum speed is:

v max = |(6.00 π rad/s) cos(0)| = 6.00 π rad/s = 18.84 cm/s

c) The maximum acceleration in simple harmonic motion occurs at the amplitude extremes (x = ±A). The acceleration can be obtained by differentiating the velocity equation:

a = d²x/dt² = -(18.00 π² rad²/s²) sin(3.00 πt)

Maximum acceleration occurs when sin(3.00 πt) = ±1, which is at t = 0.167 s, 0.833 s, 1.500 s, etc. (multiples of 1/6 cycle). Therefore, the maximum acceleration is:

a max = |-(18.00 π² rad²/s²) sin(π/2)| = 18.00 π² rad²/s² = 565.5 cm/s²

d) In one period (T = 1/f = 0.00667 s), the particle travels a distance equal to 4 times the amplitude. This is because it travels from zero to A, then to -A, back to zero, and finally to A again. To find the total distance traveled between t = 0 and t = 1.00 s, we need to consider how many complete cycles and partial cycles occur in this time:

Number of complete cycles: n = floor(1.00 s / 0.00667 s) = 150

Distance traveled in complete cycles: n * 4A = 150 * 4 * 2.00 cm = 1200 cm

Time remaining in the last incomplete cycle: 1.00 s - (n * T) = 1.00 s - (150 * 0.00667 s) = 0.00333 s

Distance traveled in the incomplete cycle: A sin(ωt) = 2.00 cm * sin(3.00 π * 0.00333 s) ≈ 0.06 cm

Therefore, the total distance traveled between t = 0 and t = 1.00 s is:

Total distance = (Complete cycles distance) + (Incomplete cycle distance) = 1200 cm + 0.06 cm ≈ 1200.06 cm

User The Disintegrator
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