SHM Book page 354 - 361 ©cgrahamphysics.com 2016

Recall

• 𝑎 ∝ to displacement from EQLB position

• 𝑎 is directed towards the EQLB position

• 𝐹 = −𝑘𝑥 = 𝑚𝑎

• 𝑎 = −𝑘

𝑚𝑥 and 𝜔 =

𝑘

𝑚= 2𝜋𝑓 =

2𝜋

𝑇

• Since T =1

𝑓=

2𝜋

𝜔= 2𝜋

𝑚

𝑘 and 𝑓 =

𝜔

2𝜋=

1

2𝜋

𝑘

𝑚

• 𝑎 = −𝜔2𝑥

©cgrahamphysics.com 2016

Circular motion and SHM

• Suppose P is rotating the perimeter of a circle

• Because rate of rotation is constant, angle 𝜃 or 𝜔𝑡 is ∝ to time

• If we draw y vs t instead, 2𝜋 𝑎𝑛𝑑 𝜋 are replaced by T and 𝑇

2

Angular speed = 𝜔 =𝜃

𝑡

𝜃 = 𝜔𝑡 𝑦 = 𝑅𝑠𝑖𝑛 𝜃 𝑥 = 𝑅𝑐𝑜𝑠 𝜃

The variation of the vertical motion with time is a sine curve

©cgrahamphysics.com 2016

Finding a solution to SHM 𝑎 = −𝜔2𝑥 • 𝑣𝑎𝑣 =

∆𝑠

∆𝑡, where s(t) is the position function

•𝑑𝑠

𝑑𝑡 is the first derivative s’(t) = v(t) = velocity

•𝑑𝑣

𝑑𝑡=

𝑑"(𝑠)

𝑑𝑡 is the acceleration a(t)

• 𝑎 𝑡 = 𝑣′ 𝑡 = 𝑠"(𝑡) =𝑑2𝑠

𝑑𝑡 is the 2nd derivative

• 𝑠 𝑡 = 𝑡2 + 2𝑡 + 12 𝑣 𝑡 = 2𝑡 + 2 𝑎 𝑡 = 2

• Hence a(t)=−𝜔2𝑥 or 𝑑2𝑥

𝑑𝑡= − 𝜔2𝑥

©cgrahamphysics.com 2016

SHM can be represented using sine or cosine curves

𝒙 = 𝒙𝟎𝒔𝒊𝒏 𝜽

• 𝑥 = 𝑥0𝑠𝑖𝑛 𝜃

• 𝑥 = 𝑥0𝑠𝑖𝑛𝜔𝑡

• 𝑣 𝑡 = 𝑥0𝜔 cos 𝜔𝑡

• 𝑎 𝑡 = −𝑥0𝜔2𝑠𝑖𝑛𝜔𝑡

𝒙 = 𝒙𝟎𝒄𝒐𝒔 𝜽

• 𝑥 = 𝑥0𝑐𝑜𝑠 𝜃

• 𝑥 = 𝑥0𝑐𝑜𝑠𝜔𝑡

• 𝑣 𝑡 = −𝑥0𝜔 sin 𝜔𝑡

• 𝑎 𝑡 = −𝑥0𝜔2𝑐𝑜𝑠𝜔𝑡

Where 𝒙𝟎 = R = A = max displacement and θ = 𝜔𝑡 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

©cgrahamphysics.com 2016

Relation between SHM and sine curve

• Sin and cos oscillate between + 1 and – 1 • Max value for 𝑥 = 𝑥0𝑐𝑜𝑠𝜔𝑡 when 𝑐𝑜𝑠𝜔𝑡 = 1

𝑥𝑚𝑎𝑥 = 𝑥0 • Max value for 𝑣 𝑡 = −𝑥0𝜔 sin 𝜔𝑡 when sin 𝜔𝑡 = 1

𝑣𝑚𝑎𝑥 = −𝑥0𝜔 • Max value for 𝑎 𝑡 = −𝑥0𝜔2𝑐𝑜𝑠𝜔𝑡 when c𝑜𝑠𝜔𝑡 = 1

𝑎0 = −𝑥0𝜔2, which is the defining equation for SHM

©cgrahamphysics.com 2016

Keeping in mind that 𝜔𝑡 varies between 0 and 2𝜋

• sin 𝜔𝑡:

• Negative for 𝜔𝑡 from 𝜋 𝑡𝑜 2𝜋

• cos 𝜔𝑡:

• Negative for 𝜔𝑡 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 ©cgrahamphysics.com 2016

Graphical representation

©cgrahamphysics.com 2016

Using trig relationships • 𝑠𝑖𝑛2𝜃 + 𝑐𝑜𝑠2𝜃 = 1

sin 𝜃 = 1 − 𝑐𝑜𝑠2𝜃

• θ = 𝜔𝑡

• sin 𝜔𝑡 = 1 − 𝑐𝑜𝑠2𝜔𝑡

• 𝑥 = 𝑥0𝑐𝑜𝑠𝜔𝑡 𝑥

𝑥0= 𝑐𝑜𝑠𝜔𝑡

• sin 𝜔𝑡 = 1 −𝑥2

𝑥02

• We know that 𝑣 𝑡 =−𝑥0𝜔 sin 𝜔𝑡

• 𝑣 𝑡 = −𝑥0𝜔 1 −𝑥2

𝑥02

= −𝜔 𝑥02 − 𝑥2

• Since v can be positive and negative, we need to write

𝑣 𝑡 = ± 𝜔 𝑥02 − 𝑥2

𝜔 = angular speed or frequency, 𝑥0 = 𝐴 = 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 𝜔 ©cgrahamphysics.com 2016

Example • An object oscillates with SHM with frequency 60Hz

and amplitude 25mm. Find the velocity at a displacement of 8mm

Solution

• 𝜔 = 2𝜋𝑓 = 2𝜋 × 60 = 120𝜋

• 𝑥0 = 25 × 10−3𝑚

• 𝑥 = 8 × 10−3𝑚

• 𝑣 𝑡 = ± 𝜔 𝑥02 − 𝑥2

= ±120𝜋 25 × 10−3 2 − 8 × 10−3 2 = ±8.9𝑚𝑠−1

©cgrahamphysics.com 2016

Boundary conditions

• When displacement 𝑥 = 𝑥0, max displacement

• At t=0 the solution is 𝑥 = 𝑥0𝑐𝑜𝑠𝜔𝑡

• An example is a simple pendulum or an harmonic oscillator

• When displacement x = 0 at EQLB

• At t=0 the solution is 𝑥 = 𝑥0𝑠𝑖𝑛𝜔𝑡

• The solutions are essential

the same, they differ by 𝜋

2

©cgrahamphysics.com 2016

General solution to SHM equation • 𝑥 = 𝑥0𝑐𝑜𝑠𝜔𝑡 + 𝑥0𝑠𝑖𝑛𝜔𝑡

• There are actually 3 solutions principle of superposition: one of the solutions to the equation is the sum of all the other solutions

• The physical quantities that 𝜔 will depend on is determined by the particular system

©cgrahamphysics.com 2016

Weight oscillating at vertical spring

• A vertical spring with spring constant k

• 𝜔 =𝑘

𝑚 and T = 2𝜋

𝑚

𝑘

©cgrahamphysics.com 2016

A simple pendulum • 𝜔 =

𝑔

𝑙 and T = 2𝜋

𝑙

𝑔

Proof • F = ma

• 𝑎 = −𝜔2𝑥 • F = −𝑚𝜔2𝑥 • Pendulum in EQLB when

T = mg cos 𝜃 • mg sin 𝜃 ≠ in EQLB and provides

restoring force F • F = mg sin 𝜃 • For small angles 𝜃, s = x

• 𝑠𝑖𝑛𝜃~𝜃 =𝑥

𝐿

mg sin 𝜃 = ma

mg 𝑥

𝐿 = ma

Restoring force: - m 𝑔

𝑙𝑥 = 𝑚𝑎 𝑎 = −

𝑔

𝑙𝑥

Hence by comparing to SHM

𝜔2 =𝑔

𝑙

©cgrahamphysics.com 2016

Example • A loudspeaker cone vibrates in SHM at a frequency of 262Hz. The

amplitude at the center of the cone is A = 1.5× 10−4𝑚 and at t = 0, x = A.

• A) what equation describes the motion of the center of the cones?

• B) what are the velocity and acceleration as a function of time?

• C) what is the position of the cone at t = 1ms?

Solution

A) 𝜔 = 2𝜋𝑓 = 2𝜋 × 262 = 1646𝑟𝑎𝑑𝑠−1 𝑥 = 𝐴 cos 𝜔𝑡 = 1.5× 10−4 cos 1646𝑡

B) 𝑣0 = 𝜔𝐴 = 1646 × 1.5× 10−4 = 0.25𝑚𝑠−1 𝑣 = −𝑣𝑚𝑎𝑥 sin 𝜔𝑡 = −0.25 sin 1646𝑡

𝑎𝑚𝑎𝑥 = 𝜔2𝐴 = 1646 2 × 1.5× 10−4 = 406𝑚𝑠−2 𝑎 = −𝑎0 cos 𝜔𝑡 = −406 cos 1646𝑡

C) 𝑥 = 𝐴 cos 𝜔𝑡 = 1.5× 10−4 cos 1646 × 1 × 10−3 = −1.3 × 10−5𝑚 ©cgrahamphysics.com 2016