New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here

Week 6 Phys 5b Notes

by: Shanee Dinay

Week 6 Phys 5b Notes PHYS 5B

Shanee Dinay
GPA 3.94

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Notes for Physics 5b Intro to Physics 2 Notes. Topics include Chapter 15 and 16, Standing Waves, Transverse Wave, String Tension, Pulse, Wave Speed, Gas Behavior, Linearization Process, Wave Intens...
Intro to Physics II
Class Notes
Physics 5b, Physics, waves, Pulse, Wave Speed, Wave Intensity
25 ?




Popular in Intro to Physics II

Popular in Physics 2

This 8 page Class Notes was uploaded by Shanee Dinay on Monday February 15, 2016. The Class Notes belongs to PHYS 5B at University of California - Santa Cruz taught by A.Steinacker in Fall 2015. Since its upload, it has received 10 views. For similar materials see Intro to Physics II in Physics 2 at University of California - Santa Cruz.


Reviews for Week 6 Phys 5b Notes


Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 02/15/16
Day 15 ­ 2/8/2016  Phys 5b    HW #1  ­ asked to calculate how long will it take for the displacement to decrease one percent to  its original value, referring to the exponential envelope. Pretend as if the sinusoidal didn’t  exist  Reading: 15.2, 15.5, 15.7, 16.1, 16.2  Review: Standing Waves  ­ by adding up two waves of the same amplitude, wavelength, and frequency, but  propagating in opposite direction, y​ 1​, t) = Asin(kx ­ wt) and y​ 2​x, t) = Asin(kx + wt), we  obtained a standing wave, and we have shown that its mode of vibration depend on the  boundary conditions as follows    Closed Closed Standing Wave Pattern  λ = 2L n = 1, 2, 3, …  n n P n = vw = n vw → T​  =  = 12L  λn 2L n​ fn nvw n↑ → T↓  → v​ p​↑  y(x, t) = Asin(kx ­ wt)  v​ =  dy = wA cos(kx ­ wt)  p​ dt   the blue is the sine graph  the purple is the velocity of the graph above it  ­ negative velocity, it is moving downward    Tension force on the rope:    forces are as follows:    An upwards pull up:      We want to use these ideas to derive the wave equation for the transverse wave:    F​ = T​  ­ T​ = T​ sin(α + aα) ­ T​ sin(α)  y​ 2y​ 1y​ 2​ 1​ T2x​= T​1x​→ T​2​= T​1   T2​os(α + aα) = T​ 1​os(α)  sin(α + a α)  → α + aα    2 dm d y = T​(α + aα) ­ T​ α  dt s​ c​ d y Ts​aα = dm dt   Linear density of our string  dm linear density =  dl  d y Ts​α = μdx  dt   2 dα = μ d2y α =  dy  d2 Tsd2 dx d y = μ d y → [T 2] =  N  =  kg • m/ =  m2  dx2 Tsdt2 μ kg /m kg / m s2 d y 1 d y =  dx2= v t2   v​ =  Ts  w​ √ μ   n = 2  vw fn​ n  2L = f  Ts​↑ 4x L = constant, f = constant  If we increase the tension, the speed of the wave is going to stay the same  v​ =  4Tso = 2v​  w0​ √ μ   w0 vw f = n 2L  n↑ → v​ w​  and we get a mode of 1    L = 2.5m  y(x, t_ = 2Acos(wt)sin(kx​ n​ xn​→ y(x​n​ t) = 0  sin(kx​) = 0  n​ kxn​= nπ  2π λ Location of mode λxn= nπ → x​ n​= n 2 λ λ λ xn+1​x​n​= (n+1) 2­ n 2=   2 λ 2  = 0.5m → λ  =  1m        Left column shows incoming pulse and ghost pulse  Right column shows us what the pulse actually looks like   as time goes on, the pulse flips with the interaction with  the wall    Keep temperature constant, then we will have direct proportion between density and pressure        Day 16 ­ 2/10/2016  Phys 5b    Waves  D(x, t) = D​m​in(kx ­ wt)  x ← D = x​ ­part​ o  D = 0 when particle pass through x​ o​for a given particle    How does P oscillate?    Speed of the Wave  Elastic prop vw​= √ Inertial prop ← ρ  String vw​=  Ts  √ μ B = − ΔP ΔP  =− B ΔV   ΔV /V V given ΔP  →  ΔV ↑ → B↓  V speed of sound c​ s​= √ ρ  gas B↓  5 kg air  Bair​ 1.5 x 10​ cs​= 343 m/s ρ  =  1.2m 3  9​ Ex. B​w​= 2.2 x 10​ Pa  ← c​w​= 1482 m/s    Take a volume of the gas V  ΔV AsΔD V = A​s​Δx ΔV = A​ sΔD ΔP  = − B V  = − B AsΔx  P​max​D​ m​k  c​ =  B   s​ √ ρ B = c​s​ w 2πf k = cs=  cs  ΔP max = 2πc sD m   Ex.   c​ = 340 m/s  s​ 3 ρ = 1.28 kg/m​   ­1 f = 2000s​   D​  = 100nm  m​ ΔP​ max​55 Pa → ( N2)  m Tube  cs fn​ n 4L n = 1, 3, 5, …  f =  cs   1​ 41 f = f​ = cs   3​ 1​ 4L1 Gas Behavior  1. P​ o​ρo vo​= 0  2. No gravity  3. small displacements & only in x­direction  4. only small variations in time & space of perturbations  ρ  =  ρo+ ρ o v = v​1  dP1 dv1 dt + ρo dx = 0  dv dv 1dP Momentum Equation: dt=   − v dx− ρ dx    Day 17 ­ 2/12/2016  Phys 5B    Linearization Process    dρ1 dv1 dt +  ρo dx = 0  dv1 dP1 ρ odx  =  ­  dt   dv1 dP1dρ0 dP 1 dP1dρ1 ρ o dt =− dρ dx dx = dρ dx   1 1 2 2 dρ1 dv1 d d o d v1 dt =  − ρ odx  ← take derivative   →  dt dt2 =   − ρodxdt  dP dρ dv dP d ρ d v − 1 0  =  ρ 1  ← take derivative  d  → − 1 2 = ρ o 1  dρ1dx o dt dx dρ1dx dxdt d ρo dP 1 ρ 1 2 1 d ρ1 add the two derivatives →  2  ­  dρ dx2  → d ρ  =o dP  / dρ 2   dt 1 1 1 dt units: [   ] =  N 1  = kg m m  = m​2 / s​  ρ m kg / m 3 s kg d ρ 1 d ρ1 dx  =  s dt     Sound Speed  2​ dP1 dP1 cs​ =  dρ1 c2​=  √ dρ1   PV = NkT k = 1.38 x 10​  ­23 P = S mk T  p dP1 k k dρ1 =  mp T → c​ s​=  mpT    √   String  T​  = T​ α → α = tanα  sy​ s T​=  ­ T​   dy sy ​ sdx P(t, x) = A​ 2T​kwcos​ 2(kx ­ wt)  s​ P  =  T​ A​2kw  2 s​ w w vw​ =  k→ k =  vw   vw​ =  T s → T​ s​= v w  √ μ 1 2 2 w 1   2 2 P =   v2μw vww =  v 2Aww   2 2 2 P  = 2π  v ρw A A   s   Wave Intensity  P 2 2 2 I  =  As 2π v ρfwA   2 P A​s​= 4πr  → I =  4πr    Inverse square law r goes up, I goes down    Interference  y 1 =  Asin(kx 1 wt  +  ϕ )01 y 2 =  Asin(kx 2 wt  +  ϕ )02 Δϕ  =  ϕ  2− ϕ  1 k(x   −2 x ) 1+  ϕ  o2 ϕ   01 Δρ  =  kΔx  +  Δϕ  →02hase length, path length difference  2π Δϕ  =   Δλ + Δϕ   o if:  Δϕ  =  2nπ perfect constructive  Δϕ  =  (2n + 1)π  perfect destructive interference   


Buy Material

Are you sure you want to buy this material for

25 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Kyle Maynard Purdue

"When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the I made $280 on my first study guide!"

Bentley McCaw University of Florida

"I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"


"Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.