Phys260 Gupta HW2 SOLUTIONS (Oscillations/Waves) REQUIRED READING Notes: So first half of this HW is still on ideas from oscillations. Problems 2 and 3 blends in ideas from oscillations with some ideas about collisions from last semester. Problems 4-6 touch upon basic ideas on waves. For Oscillations it is still Knight (14.1-14.6) or Tipler (14.1-14.3). For waves in this HW, Knight (20.1-20.3) and Tipler (15.1-15.2). 1. Consider a ball of mass 100 grams tied to a very light string from the ceiling. The length of the string is 1 meters. (This is just like a simple pendulum setup). I move the ball to the right to an angle 10 degrees and release it (at t=0) , so it starts oscillating (see figure). (a) What is the speed of the ball 0.2 seconds after I release it? [This solution contributed by Eric Kuo] I know how to find the position of the ball as a function of time. From there I can find the velocity at any time. There are two ways to think about this problem: using forces or using torques. They both lead to the same answer, but in your homework you probably went one way or other. Feel free to check the method you did on your homework, but before you read the other solution, I suggest you try it for yourself! I think the solution will do more for you if it’s helping you on something that you’ve tried yourself. Thinking about FORCES: 10 degrees is a small enough angle that we can approximate the motion of the ball as a horizontal oscillation. There are lots of ways to think about why we can approximate this. Here’s one: solve for the x and y component of the ball’s position using sin and cos. Then, use the small angle approximations: sinθ ≈ θ and cos θ ≈1. What is this approximation telling us? That the height of the ball is almost constant (L) for all small angles, but the horizontal position is changing as the angle changes (x = L θ). This is why we can approximate this as horizontal oscillation. So if the ball is only accelerating in the x-direction, then the net force in the y-direction must be zero, or else the ball would be accelerating up and down. (Ok, so in real life, the ball is actually moving up and down a little, so there is a net force in the y- direction. This is really the approximate net force that goes with our approximation of horizontal oscillation). We know that the y-component of tension of the string must be equal to mg, since it has to cancel the weight of the ball out. Using the angle of the string, we can find the x-component of the tension with a handy triangle. So the net force is in the x-direction and it’s equal to F x = –mgθ =
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Phys260 Gupta HW2 SOLUTIONS …ayush.pbworks.com/f/solutionsHW2.pdfPhys260 Gupta HW2 SOLUTIONS (Oscillations/Waves) REQUIRED READING Notes: So first half of this HW is still on ideas
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-mgx/L. Question: why is the force negative? Well, let’s say that the origin is where the angle of
the pendulum is zero. When the position of the ball is to the right, the force on it is to the left and vice
versa; that’s what a restoring force does. Mathematically, this means the sign of x and Fx must be
opposites, so we need a minus sign to make this happen.
[Alternatively, you could think about mgsin(θ) as the force along the direction of the ball’s motion
that is speeding it up as the ball moves. The other forces T-mgcosθ provide the centripetal force;
Using the approximation for small angles the net restoring force redues to mgθ – note that this force changes direction depending on whether the ball is to the right or left of the equilibrium position; The other component, always acts along the direction of tension cannot change the speed of the ball – and so can’t act as the restoring force.]
Now we have the net force, so we do what we do in oscillations: use Newton’s 2nd law to find the
position as a function of time. We know that the answer has to be cos or sin with some constants, so we
just want to figure out what those constants are. We’re thinking about it has a horizontal oscillation, so
we want to write everything in terms of x, and not θ.
The constant A is the amplitude of the oscillation. We know that the amplitude is the value of x
when θ= θ0 = 10 degrees ≈ 0.175 rad ( A = xmax = L θ0). So x(t) = L θ0 cos (√(g/L) t). The velocity of the ball
is v(t) = dx/dt = -(√(gL) θ0 sin (√(g/L) t). To find v(0.2), now just plug in all the numbers in the problem
(g=10 m/s2, L = 1m, t = 0.2 s, θ0 = 0.175 rad). The speed after 0.2 seconds is 0.035 m/s (which is positive, because the negative sign just tells you the direction of the velocity).
Thinking about TORQUES:
The torque on the ball is τ = -rFsinθ = -(L)(mg)sinθ and since 10 degrees is small enough for the
small angle approximation, τ = -Lmgθ. The torque is negative, because it always wants to rotate the
pendulum back to equilibrium. This means the torque will always have the opposite sign as the angle.
The tension of the string doesn’t cause a torque on the ball because the tension of the string is always
pulling in towards the center of the arc of the pendulum. Torques are applied to objects when the forces
on them cause them to rotate about a pivot point. Only gravity ever has a component in the direction that
the ball is moving. Mathematically, you can see that the tension causes no torque because the angle
between r and Ftension is 180 degrees and sin(180 degrees) = 0.
So we know how much torque is trying to rotate this pendulum. My motivation to find the torque
is that for forces causing oscillations in straight lines we use Newton’s 2nd law, but for rotations we want
to use the rotational version of Newton’s 2nd law, τ = Iα. The value of Iball = mL2 (if you forget why this is, I suggest checking back to your 161 notes on rotational inertia).
Now we do the usual thing that we do in oscillations: we plug everything into Newton’s 2nd law.
We know the answer has to be cos or sin with some constants, so we want to figure out those constants
using the equation we get.
The constant A is the amplitude of the oscillation. We know that the amplitude is the maximum
value of θ. This is when θ = θ0 = 10 degrees ≈ 0.175 rad. So θ (t) = θ0 cos (√(g/L) t). The angular velocity
of the ball is ω(t) = dθ/dt = -(√(g/L) θ0 sin (√(g/L) t). We also know the relationship between velocity and
angular velocity is v = Lω = -(√(gL) θ0 sin (√(g/L) t). Now, just plug in all the numbers in the problem
(g=10 m/s2, L = 1m, t = 0.2 s, θ0 = 0.175 rad). The speed after 0.2 seconds is 0.035 m/s (which is positive, because the negative sign just tells you the direction of the velocity).
NOTES:
I want to point out two things: 1) I did the problem two ways and got the same answer. It’s a good
check not just to see if you’re getting the right answer, but also to see if you really understand the ideas
behind how we are dealing with oscillations and 2) I didn’t plug in all the numbers until the end, and it’s
not just because I like working with variables (although I do). In the end, I get an expression that tells me
the velocity at any time for a pendulum at any mass with any length and initial displacement.
(b) One student has the following solution: “Well the angular frequency for the pendulum is just ω
= √(g/l) and so the speed of the ball must be v = rω. Here r = l and so v = l√(g/l) = √(gl).” What is wrong
about the student’s reasoning?
Even without going through this student’s reasoning, I can already see that the velocity cannot be
√(gl). When a pendulum swings, it speeds up on the way down and slows down on the way up. This
student is saying that the speed for the pendulum doesn’t depend on time, since g and l are constant.
So where is the mistake in this reasoning? The “ω’s” in ω = √(g/l) and v = rω do not represent the same physical idea. The first is the angular frequency of the pendulum, it is 2π divided by the period of
the pendulum. The second is the instantaneous angular velocity, how fast an object is rotating at some
particular time. If you want to know the speed of the pendulum at a particular time then it should be clear
that you want the second, the instantaneous angular velocity.
In fact, there is no way that the first ω, the angular frequency of the pendulum, can tell you the speed of the pendulum at a specific time. The angular frequency depends only on the period and the
period is just the time it takes to complete one cycle of the pendulum. This isn’t enough information to
tell you how fast the pendulum is moving at a particular time. Here’s an analogy: imagine your
roommate leaves to go to class in the physics building and comes straight back 2 hours later. Even
though you know the time it took for your roommate to leave and come back, you couldn’t say anything
about how fast he/she was walking at a particular time in the trip.
2. (These solutions adapted from Birkett
and Elby) Cart 1, of mass 1.0 kg, rolls down
the ramp shown here (height = 30 centimeters)
and collides with cart 2, of mass 2.0 kg. The
carts stick together and roll toward the spring,
which has spring constant 200 N/m and
equilibrium length 15.0 cm. Neglect friction, air resistance, and the rotational inertia of the wheels (if the
wheels are tiny and light, the rotational inertia can be quite small compared to the inertia of the cart).
(a) Taking rightward as positive, sketch rough, non-numerical graphs of position, velocity, and
acceleration vs. time for cart 2. Start your graph when cart 1 reaches the bottom of the ramp, and
end your graph when the stuck-together carts reach the bottom of the ramp after rebounding off