Respuesta :
Answer:
(a) The period is 4s
(b) The angular frequency is pi/2 radians
(c) The amplitude is 0.37cm.
(d) The displacement at time is (0.37 cm) cos((pi/2)*t)
(e) The Velocity at time t is v = (0.58 cm)(sin((pi/2)*t)
(f) The maximum speed is [tex]v_{m} = -0.58 cm/s[/tex]
(g) The maximum acceleration is 0.91 cm/s^2
Explanation:
We have a particle which oscillates with frequency of f = 0.25 Hz about the point x = 0.At t = 0, the displacement of the particle is = 0.37 cm and its velocity is zero.
(a) The period of the oscillations is,
[tex]T = 1/f[/tex]
so
T = 1/(0.25 Hz)
T = 4.0s
(b) The angular frequency is,
[tex]f = 2\pi f\\[/tex]
f = = 2([tex]\pi[/tex])(0.25 Hz) =[tex]\pi /2 \\[/tex] radians
(c) Since
The amplitude is the maximum displacement that the particle makes from the equilibrium point, or when the speed of the particle is zero,
that is
[tex]x_{m}[/tex]= 0.37 cm
(d) The displacement as a function of t is given be,
x = [tex]x_{m}[/tex] cos(ωt+Φ)
as x = [tex]x_{m[/tex] t = 0, we get cos(Ф) = 1 = 0
so this equation becomes
x= (0.37 cm) cos((pi/2)*t)
(e) Now we need to find the speed of the particle as a function of t
the speed is the derivative of the displacement that is
[tex]v = dx/dt[/tex] = -(0.37)(pi/2)(sin((pi/2)*t)
so the velocity at time t is
v = (0.58 cm)(sin((pi/2)*t)
(f) Since
[tex]v = v_{m}[/tex] sin(ωt+Ф)
then from part (e) we get
[tex]v_{m} = -0.58 cm/s[/tex]
(g)
The amplitude of the maximum acceleration is
[tex]a_{m} = x_{m[/tex] ω^2
= (0.37 cm) (pi/2) = 0.91 cm/s^2
this is the maximum acceleration
Answer:
a) Time period : T = 4 sec
b) Angular frequency : w = π / 2 rad /s
c) Amplitude : A = 0.37 cm
d) Displacement function : x(t) = 0.37*cos ( π*t / 2 )
e) Velocity function : v(t) = - 0.185*π* sin ( π*t / 2 )
f) max speed : v_max = - 0.185*π
g) max acceleration : a_max = -0.0925*π^2
Explanation:
Given:-
- The conditions for linear harmonic motion are as follows:
At t = 0 , x = 0.37 , v = 0
Mean position x = 0
Frequency f = 0.25 Hz
Find:-
Determine a) the period, (b) the angular frequency, (c) the amplitude, (d) the displacement at time t, (e) the velocity at time t, (f) the maximum speed, (g) the maximum acceleration.
Solution:-
- The time period of the SHM is the time taken (T) for one complete oscillation is as follows:
T = 1 / f
T = 1 / 0.25
T = 4 sec
- The angular frequency (w) is the speed of one complete cycle in rad/s.
w = 2*π*f
w = 2*π*0.25
w = π / 2 rad /s
- The amplitude (A) of SHM is given as the maximum displacement(x) about the mean position where velocity of the particle is zero. From given data we see that at t = 0, x = 0.37 when v = 0. Hence, A = 0.37 cm
- The displacement (x) of a particle for any time (t) undergoing SHM is given by the following expression:
x(t) = A*cos ( w*t )
x(t) = 0.37*cos ( π*t / 2 )
- The velocity (v) of a particle for any time (t) undergoing SHM can be determined by taking a time derivative of displacement relationship developed above:
v(t) = x'(t) = - 0.37*π/2 * sin ( π*t / 2 )
v(t) = - 0.185*π* sin ( π*t / 2 )
- The acceleration (a) of a particle for any time (t) undergoing SHM can be determined by taking a time derivative of velocity relationship developed above:
a(t) = v'(t) = - 0.185*π^2 / 2 * cos ( π*t / 2 )
a(t) = -0.0925*π^2* cos ( π*t / 2 )
- The maximum speed occurs when a(t) = 0,
0 = -0.0925*π^2* cos ( π*t / 2 )
0 = cos ( π*t / 2 )
t = 1 s
Hence, v = v_max at t = 1s
v_max = v(1) = -0.185*π*sin (π / 2)
= -0.185*π
- Similarly maximum acceleration occurs when t = 0,
a_max = a(0) = -0.0925*π^2*cos(0)
= -0.0925*π^2
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