q4

Helpers!! Click this link then click the links "Practice Questions" and "Practice Problems" in the left hand column.

2, 11, 12, 13, 19, 20, 21, 22, 40. For Extra Credit: 26, dealing with resonance.

Click here for 3,000 Solved Problems in Physics (Schaum's Solved Problems Series)

PLEASE REGISTER YOUR EMAIL AT THE BOTTOM OF THE PHYSICS 4A PAGE . SCROLL DOWN THAT PAGE TO FIND THE EMAIL REGISTRATION FORM !!! REGISTER AGAIN EVEN IF YOU HAVE ALREADY DONE SO; THE DATABASE HAS BEEN REFRESHED!

NOTE: In some cases the masteringphysics.com problem has been randomized with different numbers than the book problem. The hints below are done according to book problem. If odd -numbered, you can check your answer in back of book before submitting to masteringphysics.com

2.
(a) T = 33/(total number of oscillations)
(b) f = 1/T
(c) w = 6.28f
(d) A = (60 - 10)/2 in cm. Do you see why this expression is true? Why didn't I instead add the two numbers before dividing by 2?
(e) max speed = wA

11.
T = 1/f = 6.28/w. Squaring both sides we get:
T² = (6.28)²/w² , where w² =k/m. Thus:
T² = m(6.28)²/k. From this expression you can answer (a), (b) and (d). Just remember to take the positive square root.

In reference to (c), does amplitude A affect the period T?

12.
T² = m(6.28)²/k. From this expression you can find k. Note the compression distance 0.10 m does not effect your answers.

13. Much of this can be answered in a different order than presented. But let's do it in order:
(a) T = 1/f
(b) w = 6.28f
(c) Use energy conservation:
½mv₀²+ ½kx₀² = ½kA², where the values of position and velocity at t = 0 are given. Find A. Note that you need k, which can easily be found from T and mass m. See #11.
(d) See equations 14.16.
x = Acos(wt + Øo). Thus:
x(0) = AcosØo = 0.05 m
v(0) =-wAsinØo = -0.30 m/s
We see that Øo is positive, since both the sine and cosine are.
Thus we can find Øo = cos^-1(0.05/A) in radians.
(e) max speed = wA
(f) Since a_x = d²x/dt² = -w²Acos(wt + Øo), we see maximum acceleration = w²A.
(g) Total energy = ½kA²,
(h) Evaluate x = Acos(wt + Øo) at t = 0.40 seconds.

19. w² = g/L, where L = length of string. T = 1/f = 6.28/w. We see period T is independent of mass or amplitude. See example 14.8. THE AMPLITUDE INDEPENDENCE MAY NOT BE SUPRISING, SINCE THAT'S THE CASE FOR MASSES AND SPRINGS, BUT THE MISSING MASS MAY REQUIRE MORE DISCUSSION. Why doesn't bob mass m appear? The short answer is that it canceled out.

20. Please be careful. Make sure your calculator has the correct mode., either degree (deg) or radians (rad). For example, if you compute cos 30, entering 30 while in rad mode, you will get the wrong answer. Likewise if you compute cos 3.14, entering 3.14 while in deg mode you will be wrong.

(a) See the expression for θ(t) to see the amplitude.
(b) ω = 5, so find the frequency f in Hz.
(c) See the expression for θ(t) to see the phase constant.
(d) ω² = g/L
(e) Using formula, set t = 0 to see the initial angle.
(f) Using formula, set t = 2.0 seconds to evaluate angle at that time.

21. θ(t) = Acosωt, where ω² = g/L. Get the period T from the given information. Use the relationship between ω and T to get the length L. Note A and the mass were not needed.

22. ω² = g/L. Use the relationship between ω and T. Then find g for the Earth and Venus . Earth g is well known; we've been using it all semester, to four sig. fig, in the lab. Venus g is found using g = GM/R² , where M and R are the mass and radius of the planet . That data is in the astronomical table at the back of book.

40.
(a) See hint to 13, part (c) The good news is that k is given explicitly.
(b) Evaluate equation 14.29 when x = -A. See comment below.
(c) The acceleration has maximum magnitude when x = +A or -A. The maximum acceleration occurs at maximum compression, corresponding to x = -A. That's because the force at that position points in the positive x-direction.
(d) Use ½mv_x²+ ½kx² = ½kA² to find | v_x| at the position given.

More hints posted soon !