Which of the following is necessary to make an object oscillate?
i. a stable equilibrium ii. little or no friction iii. a disturbance
A: i only B: ii only C: iii only D: i and iii E: All three
Answer: All three are generally necessary. If you don't have a stable equilibrium, the object won't oscillate, it'll "run away". If you have too much friction, oscillations are damped, and don't look
"oscillatory" any more. (Think of a car with good shocks. It doesn't bounce up and down - it just goes down once, and comes back up). For most objects, you need a disturbance to take you OUT of
equilibrium, so you can oscillate AROUND equilibrium. (But I suppose you could argue an electron oscillates in an atom without the need for any disturbance?)
at equilibrium at the origin. When displaced from the origin, the particle experiences a restoring force proportional to the displacement from the origin.
Which of these equations represents this situation?
(in these equations, k is a positive constant)
A) F = – k x B) F = –k
C) F = x – k D) F = –x + k E) None of these.
Answer: F=-k x is correct. It's "proportional to displacement", the means it's proportional to x, which mean it looks like a constant times x. "Restoring" implies the force is *towards* the origin, i.e. you need a minus sign.
x O
A mass is oscillating back and forth on a spring without
friction, as shown. At which position is the magnitude of the acceleration of the mass a maximum? Position 0 is the
relaxed (unstretched) position of the mass.
0 M E
A: 0 B: M C: E
Answer: E, the place where it is MOST stretched, provides the MOST force.
At what point is the total energy of the system a maximum?
A: 0 B: M C: E
D: None of these, total energy is constant
Answer: D is completely correct! Springs conserve energy, it is merely shifting back and forth between kinetic energy and potential energy. At point "O" the KE is maximized, at point E the PE is maximized, but at ALL points the total energy is always the same.
The position of a mass on a spring as a function of time is shown below. At the time corresponding to point P,
A: The velocity v > 0 and acceleration a < 0 B: v < 0 and a > 0
C: v > 0 and a > 0 D: v < 0 and a < 0
Hint: Where does point P move to a short time later?
Answer: A short time later, x has INCREASES a bit, which means v>0. However, we are PAST the
"equilibrium" point (x=0), and nearly at our max, so the force is already pulling us back towards the origin, i.e. F<0, which means a <0.
Or, mathematically, dx/dt is positive, but d^2x/dt^2 is negative (it's concave down there)
x
t
P
The solid curve is a graph of x t( )= Acos( )wt .
The dotted curve is a graph of x t( )=Acos(wt+f) where f is a phase constant whose magnitude is less than p/2.
Is f positive, negative or zero?
A: Positive B: Negative. C: Zero
[Hint: cos(q) reaches a maximum when q=0, that is, at cos(0).]
Answer: You need to really think about this yourself! I claim it's negative. Consider some SMALL positive time t just after zero when the dashed curve has reached the top. So right there, the argument of the cos must be zero. That means +wt + phi = 0 for some small positive t. That means phi must be some small NEGATIVE value!
x
t
A
-A
A mass on a spring oscillates with a certain amplitude and a certain period T. If the mass is doubled, the spring constant of the spring is doubled, and the amplitude of motion is doubled, the period ..
A: increases B: decreases
C: stays the same.
D: Not enough information to decide
Answer: You got to think carefully about this in tutorial. Just mathematically, T = 2 pi Sqrt[m/k]
So if you double m AND k, T stays exactly the same. The amplitude has *nothing* to do with it!
CT16-7 A particle-on-a-spring oscillates with an amplitude A, and a period T.
ANSWERS:
A: twice as big.
B: 4 times as big.
C: 1/2 as big.
D: 1/4 as big.
E: Stays the same
If you double the amplitude, the maximum velocity becomes
Answer: Energy goes like 1/2 k A^2. SO if you double A, you quadruple the energy. But the max velocity is also 1/2 m v^2. SO quadruple the energy means double the velocity.
Another way to think about this is that if x = A sin(omega*t), then v = dx/dt = A*omega*cos(omega*t)
So if you double A, (without changing omega), then you must double v as well. Same answer.
If you double the amplitude, the energy of the system becomes
Answer: 4 times bigger, as stated above.
If you double the amplitude, the maximum acceleration becomes
Answer:F = -kx, so if you double x(max) (which is the same thing as "amplitude", then you have also doubled the maximum force, and that means (since F=ma) you've doubled the max acceleration.
Or you can use a = dv/dt = -A omega^2*sin(omega*t), which is proportional to A.
If you double the amplitude, the period becomes
Answer:No change, period is independent of amplitude.
An object hangs motionless from a spring. When the object is pulled down and held, the sum of the potential energy of the spring and the gravitational potential energy of the object A: increases.
B: stays the same.
C: decreases.
Answer:
Since you pulled it, you did some work on it, which means the energy of the system has gone UP.
That's it, that's the answer.
Before, it was sitting at rest. You could, if you wanted, call that point "zero energy". (Remember, you can always call any point you want the zero of energy). Now you pull it, and it *manifestly* has some more energy, because if you let go it starts bobbing up and down.
If you think about it as U(spring)+U(grav) = 1/2k x^2 + mgh, then you might get confused. Because the first term went UP, but the second term went DOWN. So who wins? (Well... the one that's squared!)
The force on a pendulum mass along the direction of motion is mgsinq.
For small q, mgsinqªmgq, and the period is independent of
amplitude. For larger amplitude motion, the period
A: increases B: decreases
C: remains constant
Hint: does sinq get bigger or smaller than q as q increases.
Answer: sin(theta) gets bigger... but not as fast as theta does! (After all, theta can keep on increasing, but sin(theta) ultimately has to stop at +1. So sin(theta) is always at least a little SMALLER than theta. So the force is a little SMALLER than we approximated. Smaller force means a little LONGER time. So the period should get a little LARGER than we thought if the amplitude is large.
mgsin mg
A kid is swinging on a swing with a period T. A second kid climbs on with the first, doubling the weight on the swing.
The period of the swing is now...
A: the same, T B: 2T C: T Sqrt[2]
D: T/Sqrt[2] E: None of these
Answer: For a pendulum, T = 2 pi Sqrt[l/g]. Adding weight doesn't impact the period.
(The tutorial on SHM should have helped you understand WHY that is)
The period of a physical pendulum is T= 2p mgLI . Compare the periods of two physical pendula. One is a solid disk of mass m, radius R, supported at the edge. The other is a hoop also of mass m, radius R, supported at the edge.
Which has the longer period?
A: Disk B: Hoop C: The periods are the same.
Answer: Hoop has I=MR^2 about it's CM, Disk has I=(1/2) MR^2. Then you have to use the parallel axis theorem to add MR^2 to each of those to get I about the pivot point. Apparently, the I of the hoop is *larger*, and therefore T is longer.
On the moon, is the period different than on the Earth?
A: longer on Moon B: shorter C: Periods are the same.
Answer: g is smaller, so T is longer. Makes sense, in outer space, you'd have an infinite period, because there's not gravity at all pulling you back to equilibrium. On the moon, the pull is about 6 times weaker, so Sqrt[6] is between 2 and 3 times longer swing period. Don't bring your grandfather clock when you go to work on Luna-Base!
pivot
C.M.
pivot
C.M.
Disk Hoop
The period of a physical pendulum is T= 2p mgL . What happens to the period T of a "hoop" physical
pendulum, (see last question) when the mass is doubled?
(Careful! What happens to I?) A: T(new)=T(old)
B: T(new)= T(old)/2
C: T(new)=T(old)/Sqrt[2]
Answer: I = (MR^2+MR^2) = 2 MR^2. If you double the mass, both numerator and denominator double, so NOTHING happens to the period. (True for a simple pendulum also, with
T = 2 pi Sqrt[l/g] )
A person swings on a swing. When the person sits still, the swing oscillates back and forth at its natural frequency. If, instead, the person stands on the swing, the natural
frequency of the swing is..
A: greater. B: the same. C: smaller.
(Hint: When the person stands, the pendulum is changing from approximately a simple pendulum to approximately a physical pendulum.)
Answer: Intuitively, if you stand, it seems like you have effectively a shorter pendulum, which has a shorter period, or HIGHER frequency.
If you prefer to use the formula, T = 2 pi Sqrt[ I/ m L g], then I has gone from mL^2 (which is what you had when the person sits, and all the mass was out a distance L away from the pivot) to I = (fraction)*mL^2, because the standing person has shifted some mass to smaller radii.
So the period gets shorter - higher frequency.
A stiff spring and a floppy spring have potential energy diagrams shown below. Which is the stiff spring?
A B
Answer: Stiff spring means large k, it means for a given amount of x, you should have LOTS of kx^2.
So B must be the stiffer spring- the PE gets larger faster.
Two masses are identical. One is attached to a stiff spring;
the other to a floppy spring. Both are positioned at x=0 and given the same initial speeds. Which spring produced the largest amplitude motion?
A: The stiff spring B: The floppy spring
Answer: If both have the same initial speed, both systems begin with the same total energy. Using the curves above, drawing a horizontal line (the total energy) at some height, I see that the floppy curve A will intersect that line at LARGER X, so the floppy spring will produce larger swing of motion.
This makes physical sense - a floppy spring will need to stretch farther before you can slow that thing down!
Pink
x PE
Yellow
x PE
