SHM Book page 354 - 361 - · PDF fileGraphical representation ©cgrahamphysics.com 2016...
Transcript of SHM Book page 354 - 361 - · PDF fileGraphical representation ©cgrahamphysics.com 2016...
SHM Book page 354 - 361 Β©cgrahamphysics.com 2016
Recall
β’ π β to displacement from EQLB position
β’ π is directed towards the EQLB position
β’ πΉ = βππ₯ = ππ
β’ π = βπ
ππ₯ and π =
π
π= 2ππ =
2π
π
β’ Since T =1
π=
2π
π= 2π
π
π and π =
π
2π=
1
2π
π
π
β’ π = βπ2π₯
Β©cgrahamphysics.com 2016
Circular motion and SHM
β’ Suppose P is rotating the perimeter of a circle
β’ Because rate of rotation is constant, angle π or ππ‘ is β to time
β’ If we draw y vs t instead, 2π πππ π are replaced by T and π
2
Angular speed = π =π
π‘
π = ππ‘ π¦ = π π ππ π π₯ = π πππ π
The variation of the vertical motion with time is a sine curve
Β©cgrahamphysics.com 2016
Finding a solution to SHM π = βπ2π₯ β’ π£ππ£ =
βπ
βπ‘, where s(t) is the position function
β’ππ
ππ‘ is the first derivative sβ(t) = v(t) = velocity
β’ππ£
ππ‘=
π"(π )
ππ‘ is the acceleration a(t)
β’ π π‘ = π£β² π‘ = π "(π‘) =π2π
ππ‘ is the 2nd derivative
β’ π π‘ = π‘2 + 2π‘ + 12 π£ π‘ = 2π‘ + 2 π π‘ = 2
β’ Hence a(t)=βπ2π₯ or π2π₯
ππ‘= β π2π₯
Β©cgrahamphysics.com 2016
SHM can be represented using sine or cosine curves
π = πππππ π½
β’ π₯ = π₯0π ππ π
β’ π₯ = π₯0π ππππ‘
β’ π£ π‘ = π₯0π cos ππ‘
β’ π π‘ = βπ₯0π2π ππππ‘
π = πππππ π½
β’ π₯ = π₯0πππ π
β’ π₯ = π₯0πππ ππ‘
β’ π£ π‘ = βπ₯0π sin ππ‘
β’ π π‘ = βπ₯0π2πππ ππ‘
Where ππ = R = A = max displacement and ΞΈ = ππ‘ = ππππ’πππ π£ππππππ‘π¦
Β©cgrahamphysics.com 2016
Relation between SHM and sine curve
β’ Sin and cos oscillate between + 1 and β 1 β’ Max value for π₯ = π₯0πππ ππ‘ when πππ ππ‘ = 1
π₯πππ₯ = π₯0 β’ Max value for π£ π‘ = βπ₯0π sin ππ‘ when sin ππ‘ = 1
π£πππ₯ = βπ₯0π β’ Max value for π π‘ = βπ₯0π2πππ ππ‘ when cππ ππ‘ = 1
π0 = βπ₯0π2, which is the defining equation for SHM
Β©cgrahamphysics.com 2016
Keeping in mind that ππ‘ varies between 0 and 2π
β’ sin ππ‘:
β’ Negative for ππ‘ from π π‘π 2π
β’ cos ππ‘:
β’ Negative for ππ‘ from π
2
to 3π
2
This means
When displacement from EQLB is positive
then v is negative and so directed towards EQLB
This means
When displacement from EQLB is negative
then v is positive and so directed away from EQLB Β©cgrahamphysics.com 2016
Graphical representation
Β©cgrahamphysics.com 2016
Using trig relationships β’ π ππ2π + πππ 2π = 1
sin π = 1 β πππ 2π
β’ ΞΈ = ππ‘
β’ sin ππ‘ = 1 β πππ 2ππ‘
β’ π₯ = π₯0πππ ππ‘ π₯
π₯0= πππ ππ‘
β’ sin ππ‘ = 1 βπ₯2
π₯02
β’ We know that π£ π‘ =βπ₯0π sin ππ‘
β’ π£ π‘ = βπ₯0π 1 βπ₯2
π₯02
= βπ π₯02 β π₯2
β’ Since v can be positive and negative, we need to write
π£ π‘ = Β± π π₯02 β π₯2
π = angular speed or frequency, π₯0 = π΄ = max displacement, x = position at specific instant This equation is useful for finding velocity at any particular position when knowing amplitude and period or f or π Β©cgrahamphysics.com 2016
Example β’ An object oscillates with SHM with frequency 60Hz
and amplitude 25mm. Find the velocity at a displacement of 8mm
Solution
β’ π = 2ππ = 2π Γ 60 = 120π
β’ π₯0 = 25 Γ 10β3π
β’ π₯ = 8 Γ 10β3π
β’ π£ π‘ = Β± π π₯02 β π₯2
= Β±120π 25 Γ 10β3 2 β 8 Γ 10β3 2 = Β±8.9ππ β1
Β©cgrahamphysics.com 2016
Boundary conditions
β’ When displacement π₯ = π₯0, max displacement
β’ At t=0 the solution is π₯ = π₯0πππ ππ‘
β’ An example is a simple pendulum or an harmonic oscillator
β’ When displacement x = 0 at EQLB
β’ At t=0 the solution is π₯ = π₯0π ππππ‘
β’ The solutions are essential
the same, they differ by π
2
Β©cgrahamphysics.com 2016
General solution to SHM equation β’ π₯ = π₯0πππ ππ‘ + π₯0π ππππ‘
β’ There are actually 3 solutions principle of superposition: one of the solutions to the equation is the sum of all the other solutions
β’ The physical quantities that π will depend on is determined by the particular system
Β©cgrahamphysics.com 2016
Weight oscillating at vertical spring
β’ A vertical spring with spring constant k
β’ π =π
π and T = 2π
π
π
Β©cgrahamphysics.com 2016
A simple pendulum β’ π =
π
π and T = 2π
π
π
Proof β’ F = ma
β’ π = βπ2π₯ β’ F = βππ2π₯ β’ Pendulum in EQLB when
T = mg cos π β’ mg sin π β in EQLB and provides
restoring force F β’ F = mg sin π β’ For small angles π, s = x
β’ π πππ~π =π₯
πΏ
mg sin π = ma
mg π₯
πΏ = ma
Restoring force: - m π
ππ₯ = ππ π = β
π
ππ₯
Hence by comparing to SHM
π2 =π
π
Β©cgrahamphysics.com 2016
Example β’ A loudspeaker cone vibrates in SHM at a frequency of 262Hz. The
amplitude at the center of the cone is A = 1.5Γ 10β4π and at t = 0, x = A.
β’ A) what equation describes the motion of the center of the cones?
β’ B) what are the velocity and acceleration as a function of time?
β’ C) what is the position of the cone at t = 1ms?
Solution
A) π = 2ππ = 2π Γ 262 = 1646ππππ β1 π₯ = π΄ cos ππ‘ = 1.5Γ 10β4 cos 1646π‘
B) π£0 = ππ΄ = 1646 Γ 1.5Γ 10β4 = 0.25ππ β1 π£ = βπ£πππ₯ sin ππ‘ = β0.25 sin 1646π‘
ππππ₯ = π2π΄ = 1646 2 Γ 1.5Γ 10β4 = 406ππ β2 π = βπ0 cos ππ‘ = β406 cos 1646π‘
C) π₯ = π΄ cos ππ‘ = 1.5Γ 10β4 cos 1646 Γ 1 Γ 10β3 = β1.3 Γ 10β5π Β©cgrahamphysics.com 2016