A particle executes shm of period 5s it's velocity is found to be 4 cm/ s after 0.8s it has crossed the mean position Calculate it's velocity when the displacement is 2.5m
Answers
Explanation:
where distance is measured in metres and time in seconds.
(a) What is the amplitude, frequency, angular frequency, and period of this motion?
(b) What is the equation of the velocity of this particle?
(c) What is the equation of the acceleration of this particle?
(d) What is the spring constant?
(e) At what next time t > 0, will the object be:
i. at equilibrium and moving to the right,
ii. at equilibrium and moving to the left,
iii. at maximum amplitude, and
iv. at minimum amplitude.
(a) First write the general expression on top of the given expression
x = Acos(ωt + ϕ0
)
x = 4cos(1.33t + π/5)
Immediately, we get A = 4 m, ω = 1.33 rad/s, and ϕ0 = π/5. Since ω = 2π / T = 2πf, we get T = 2π /1.33 rad/s = 4.724 s,
and f = ω / 2π = 0.2117 s
−1
.
(b) The velocity is given by the first derivative of position with respect to time
v = −ωAsin(ωt + ϕ0
) .
With the given values, we get
v = −5.32sin(1.33t + π/5) .
(c) The acceleration is given by the second of position with respect to time, or the first derivative of the velocity with respect
to time,
a = −ω
2A cos(ωt + π/5) .
With the given values, we get
a = −7.08cos(1.33t + π/5) .
(d) We have the relation that ω
2 = K/m, so
K = ω
2m = (1.33 rad/s)
2
(1.75 kg) = 3.0956 N/m .
(e) (i) & (ii) We know that at equilibrium x = 0. We also know that there are two places where this happens, one where is the
velocity is positive and the object is moving to the right, and one where the velocity is negative and the object is moving to the
left. So first let's set x = 0,
0 = 4cos(ωt + π/5) .
We can divide through by 4, and we get
0 = cos(ωt + π/5) .
Taking the inverse of both sides, the solution is
ωt + π/5 = cos
−1
(0) ,
and thus,
t = [cos
−1
(0) − π/5] / ω .
Now cos
−1
(0) has many solutions, all the angles in radians for which the cosine is zero. This occurs for angles θ = π/2, θ =
−π/2, θ = 3π/2, θ = −3π/2, and so on. This is usually expressed
θ = nπ/2, where n = ±1, ±3, ±5, …
So our solutions for t are in the form
t = [nπ/2 − π/5] / ω, where n = ±1, ±3, ±5, …
The first nonzero time when x = 0 occurs for n = +1,
t1 = [π/2 − π/5] / = 0.7086 s.
The second nonzero time occurs when n = +2,
t2 = [3π/2 − π/5] / = 3.0707 s.
To tell which has the object moving to the right and which to the left we examine the velocity
v1 = −5.32sin(1.33t1 + π/5) = −5.32sin(π/2) = −5.32 m/s,
v2 = −5.32sin(1.33t2 + π/5) = −5.32sin(3π/2) = +5.32 m/s.
We see that the object is moving to the left, has negative velocity, at t = t1 = 0.7086 s, and is moving to the right at t = t2 =
3.0707 s.
(iii) At maximum amplitude, x = +4, so we have
4 = 4cos(ωt + π/5) .
Dividing through by 4, we get
1 = cos(ωt + π/5)
Taking the inverse of both sides, the solution is
ωt + π/5 = cos
−1
(1) ,
and thus,