simple harmonic motion1
DESCRIPTION
TRANSCRIPT
a – x
a = – 2x
Negative sign:Direction of a is opposite to x
O-xoxo
a a
200grams
VibratingTuning fork
A weight ona spring
A boy on a swing
F – x
F = – kx
O-xoxo
Point of equilibrium:a = 0F = 0
a
x
Measured from the point of equilibrium
Amplitude := xo
=Maximum distance
F = - kx
ma = - kx
a = km
- x a = - 2xcompare
2 =km
= angular frequency
= 2f = 2 T
F
x0 xo-xo
a
x0 xo-xo
2xo
-2xo
F = -kx a = - 2x
a = - 2x
d2xdt2 = - 2x
x = xo sin (t + )General solution
dxdt
= xo cos (t + )v =
d2xdt2
= -2xo sin (t + )a =
= - 2x
x = xo sin (t + )
phase
Phase constant
x = xo sin (t + ) depends on the initial condition
when t = 0
O-xoxo
At t = 0, x = 0
0 = xo sin ((0) + )
sin (0 + ) = 0
sin = 0
= 0
Therefore :
x = xo sin t
v = xo cos t
a = - 2xo sin t
= - 2x
x
t0
x = xo sin txo
-xoT 2T
v
T0-xo
xov = xo cos t
t2T
a
t0
a = -2xo sin t2xo
-2xoT 2T
t=0,v=xo
t=0,a=0
phase = t + /2
phase = t
x
t0
xo
-xoT 2T
v
T0-xo
xo
t2T
a
t0
2xo
-2xoT 2T
x = xo sin t
v = xo cos t
To compare the phase :They must all be expressed
as same function (either as sine
or cosine function)
Phase difference :=(t+ ) - t
2 =
2rad
= xo sin (t + /2)
v leading x
phase = t +
x
t0
xo
-xoT 2T
v
T0-xo
xo
t2T
a
t0
2xo
-2xoT 2T
x = xo sin tphase = t
v = xo cos t phase = t + /2
a = -2xo sin t = 2xo sin (t + )
Phase difference :=(t+) - t = rad
a leading xa and x are antiphase
x
t0
xo
-xoT 2T
v
T0-xo
xo
t2T
a
t0
2xo
-2xoT 2T
-xo xo0
t=0
x
x = xo sin (t + )General solution
Substitute :t=0 , x = xo
to find xo = xo sin ((0) + )
sin = 1
=2
Therefore , equation of x is: x = xo sin (t + /2)
t
= xo cos t
The motion of a body in simple harmonic motion is described by the equation :
x = 4.0 cos ( 2t + )3
Where x is in metres, t is in seconds and (2t + ) is in radians.
3a)What is
i) the displacement ii) the velocity iii) the acceleration and iv) the phaseWhen t = 2.0 s ?
b) Find i) the frequency ii) the periodof the motion.
x = 4.0 cos ( 2t + )3a) i)
Displacement = ?
t = 2.0 s ; x = 4.0 cos ( 2(2) + )3
= 2.0 m
ii) velocity = ?
v = dxdt
= -2(4.0) sin ( 2t + )3
v = -8.0 sin (4 + )t = 2.0 s ;3
= -21.8 ms-1
iii) acceleration = ?
a = dvdt
= -16.02 cos ( 2t + )3
t = 2.0 s ; a = -16.02 cos ( 2(2) + )3
= -79.0 ms-2
iv) Phase = ?
x = 4.0 cos ( 2t + )3
Phase = 2t + 3
t = 2.0 s ; Phase = 2(2) + 3
= 13.6 rad
The motion of a body in simple harmonic motion is described by the equation :
x = 4.0 cos ( 2t + )3
Where x is in metres, t is in seconds and (2t + ) is in radians.
3a)What is
i) the displacement ii) the velocity iii) the acceleration and iv) the phaseWhen t = 2.0 s ?
b) Find i) the frequency ii) the periodof the motion.
x = 4.0 cos ( 2t + )3
b) i) frequency = ?
x = xo cos (t + )
General solution :
= 2
2f = 2
f = 1.0 Hz
ii) Period = ?
T = 1 f
= 1.0 s
Figures (1) and (2) are the displacement-time graph and acceleration-time graph respectively of a body in simple harmonic motion.
x,m3
-3
t,sT0
(1) a,ms-2
12
-12
t,sT0
(2)
What is the frequency of the motion? Write an expression to represent the variation of displacement x with time t.
x,m3
-3
t,sT0
(1) a,ms-2
12
-12
t,sT0
(2)
frequency = ?
xo = 3 2xo = 12
2(3) = 12substitute
= 2 rad s-1
2f = 2f = 0.318 Hz
expression = ?
x = xo sin (t + )General solution
xo = 3 , = 2 ; x = 3 sin (2t + )
At t = 0, x = 3 3 = 3 sin (2(0) + )
sin = 1
= 2
rad
Therefore, the expression is
x = 3 sin (2t + )2
x = 3 cos 2t
a = dv = -2x dt
dv x dx = -2xdx dt
v dvdx
= -2x
v dv = -2x dx
v2
2= -2x2
2+ C
When x = xo, v = 0
0 = -2xo2
2+ C
C = 2xo2
2Hence ;
v2
2= -2x2
2 2xo
2 2
+
v2 = -2x2 + 2xo2
v = xo2 – x2
v
x0
-xo O xo
xo
-xo
xo-xo
v = xo2 – x2
A particle performs simple harmonic motion with an amplitude of 0.50 cm and a period of 3.0 s. Calculate
a)The angular frequency of the SHMb)The velocities of the particle when its
displacement is 0.20 cm. Explain why there are two velocities.
a) = 2f
= 2 T
= 2 3
= 2.09 rad s-1
b) v = xo2 – x2
= (2.09) 0.52 – 0.22
= 0.958 cm s-1
v = +0.958 cms-1 when the particle is moving in the positive direction and v = –0.958 cms-1 when the particle passes through the same point in the opposite direction on the return journey.
Total energy = E = K + U
At point, x = 0 :
K = ½ mv2
= ½ m2(xo2 – x2)
K = ½ m2(xo2 – 0)
= ½ m2xo2
At point, x = 0 :
U = 0
= maximum
= ½ m2xo2 + 0
= ½ m2xo2
constant
At point, x = xo :
K = ½ m2(xo2–xo
2) = 0
= minimum
-xo xo0 x
-xo xo0 x
At any displacement,x
E = K + U
½ m2xo2 = ½ m2(xo
2 – x2) + U
= ½ m2xo2 – ½ m2x2 + U
U = ½ m2x2
Kmax = ½ m2xo2 U = ½ m2x2Kmin = 0; ;
0
E
x-xo xo
Etotal=½ m2xo2
-xo xo0 x
K=½ m2(xo2 – x2)
U = ½ m2x2
-xo xo0 x
U = ½ m2x2
F = – dU dx
= – ½ (2) m2x
= – m2x
constant
Therefore: F x
0
F
x-xo xo
mxo
-mxo
-xo xo0 x
x = xo sin (t + )General solution
0 = xo sin ((0) + )At t = 0 , x = 0
0 = sin
= 0
x = xo sin t Therefore :The equation is
-xo xo0 x
x = xo sin t v = dx dt
= xo cos t
K = ½ mv2 = ½ m2xo2cos2 t
U = ½ m2x2 = ½ m2xo2 sin2 t
Etotal=½ m2xo2
T 2T
K
0
E
tU