GSI Aaron Practice Final (solutions)

GSI Aaron Practice Final (solutions) - Practice Final...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Practice Final Solutions Physics 7A, UC Berkeley, Spring 2011 Scoring Problem Max Points Your score 1(A) 4 1(B) 4 2 8 3 4 4(A) 3 4(B) 3 4(C) 2 5 4 6(A) 4 6(B) 2 6(C) 2 Total 40 Problem 1(A). This is a resonance question: the engineer designs the road so that if the car drives at a certain speed, the resonance will cause the car to vibrate wildly, forcing the driver to slow down. We have to see what the resonant (in this case, natural) frequency of the car is. The four tires act as four springs in parallel; thus, their respective forces are added together. It as if one spring of 1 . 4 × 10 6 N/m were acting on the car. Thus, the natural frequency of the car is ω n = ° k m = ± 1 . 4 × 10 6 N/m 1200 kg =34 . 16 rad/s (1) Now we have to establish the frequency with which the road acts on the car. The speed bump acts like a sinusoidal force acting on the tires. A speed bump is one half (just the part from 0 to π )o fas inewave . Thus ,the“wave length”isactua l ly1 . 6 m .W ek n owt h e resonant frequency of the car and the wavelength of the road. At what velocity, then, will the driver experience this resonance? We know the relationship v = λf = λ ω n 2 π .W ecanju s t plug in our values. v = λf = λ ω n 2 π =(1 . 6 m ) ² 34 . 16 s
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
response—in this case, how much the car will “bump”—is very low. Scoring: 1po intfor ω n = ° 4 k m ,2po intsforre lat ingve loc ityandwave lengthtothedr iv ing force frequency, 1 point for explanation of why the professor speeds up = 4 points. Figure 1: Cart-Mass Oscillatory System for Problem 1(B). Problem 1(B). Let us set an x-y axis as indicated in ±gure 1, such that the x axis is on the cart’s track and the y axis is right above the center of mass. The only external forces acting on the system are gravity and the normal, both forces in the y direction. Thus, although the CM may move up and down, it will stay in the same x position, or x =0inourcoord inate system. If we use ζ as the length along the string and set the zero at the cart, then the ball is located at ζ =0 . 4 m .Thu s ,thec en te ro fma s si sloca teda t ¯ ζ = (0)(3 kg )+( . 4 m )(1 kg ) 4 kg = . 1 m. (3) The CM is 10 cm from the cart along the string. Thus, the position of the cart is r cart = . 1sin θ i and the position of the CM is ¯r = . 1cos θ j .Thepo s i t iono ftheba l lre la t iv etothe CM is r ball , rel = . 3sin θ i . 3cos θ
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 7

GSI Aaron Practice Final (solutions) - Practice Final...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online