The equation of motion of a particle in vertical SHM is given by y = (10 cm) sin 0.80t.
(a) What is the particle's displacement at t = 1.1 s?
(b) What is the particle's velocity at t = 1.1 s?
(c) What is the particle's acceleration at t = 1.1 s?
remember !! 0.8 rad
y = 10•sin 0.8t = 10•sin 0.8•1.1 =7.7 cm
y`= 10•0.8•cos0.8•1.1 =5.1 cm/s,
y`` = 10• (0.8)^2• sin 0.8•1.1 =
= 4.9 cm/s^2
To find the answers to these questions, we will use the given equation of motion: y = (10 cm) sin(0.80t).
(a) To find the particle's displacement at t = 1.1 s, substitute t = 1.1 s into the equation:
y = (10 cm) sin(0.80 * 1.1)
= (10 cm) sin(0.88)
≈ 8.69 cm
Therefore, the particle's displacement at t = 1.1 s is approximately 8.69 cm.
(b) To find the particle's velocity at t = 1.1 s, we need to find the derivative of the given equation with respect to time:
v = dy/dt = (10 cm) * (0.80) * cos(0.80t)
= 8.0 * cos(0.80t)
Substitute t = 1.1 s into the equation:
v = 8.0 * cos(0.80 * 1.1)
≈ 6.82 cm/s
Therefore, the particle's velocity at t = 1.1 s is approximately 6.82 cm/s.
(c) To find the particle's acceleration at t = 1.1 s, we need to find the second derivative of the given equation with respect to time:
a = dv/dt = d^2y/dt^2 = (10 cm) * (0.80)^2 * (-sin(0.80t))
= -6.4 * sin(0.80t)
Substitute t = 1.1 s into the equation:
a = -6.4 * sin(0.80 * 1.1)
≈ -5.45 cm/s^2
Therefore, the particle's acceleration at t = 1.1 s is approximately -5.45 cm/s^2.
To find the particle's displacement at t = 1.1 s, we can plug in the value into the equation y = (10 cm) sin(0.80t):
(a) Displacement at t = 1.1 s:
y = (10 cm) sin(0.80 * 1.1)
y ≈ (10 cm) sin(0.88)
y ≈ (10 cm) * 0.7689
y ≈ 7.689 cm
Therefore, the particle's displacement at t = 1.1 s is approximately 7.689 cm.
To find the particle's velocity at t = 1.1 s, we need to first find the derivative of the displacement equation with respect to time:
v(t) = d/dt [(10 cm) sin(0.80t)]
(b) Velocity at t = 1.1 s:
v(t) = (10 cm) * d/dt [sin(0.80t)]
v(t) = (10 cm) * (0.8) * cos(0.80t)
Now we can calculate the velocity at t = 1.1 s:
v(1.1) = (10 cm) * (0.8) * cos(0.80 * 1.1)
v(1.1) ≈ (10 cm) * (0.8) * cos(0.88)
v(1.1) ≈ (10 cm) * 0.8 * 0.6428
v(1.1) ≈ 5.1424 cm/s
Therefore, the particle's velocity at t = 1.1 s is approximately 5.1424 cm/s.
To find the particle's acceleration at t = 1.1 s, we need to find the second derivative of the displacement equation:
a(t) = d²/dt² [(10 cm) sin(0.80t)]
a(t) = (10 cm) * d²/dt² [sin(0.80t)]
a(t) = (10 cm) * (0.8)² * (-sin(0.80t))
(c) Acceleration at t = 1.1 s:
a(t) = (10 cm) * (0.8)² * (-sin(0.80 * 1.1))
a(t) ≈ (10 cm) * (0.8)² * (-sin(0.88))
a(t) ≈ (10 cm) * (0.8)² * (-0.6116)
a(t) ≈ -3.0558 cm/s²
Therefore, the particle's acceleration at t = 1.1 s is approximately -3.0558 cm/s².