course phy 231
036. `query 36
Question: **** `qQuery class notes #37
If we know the angular frequency `omega and the amplitude A of motion how do we obtain an equation of motion (i.e., the formula that gives us the position of the pendulum if we know the clock time t)? What are the corresponding velocity and acceleration functions?
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Your solution:
position is Acos(wt+theta)
veloctity is -wasin(wt+theta)
acceleration is -w^2Acos(wt+theta)
the thetas should be the signol for a starting theta position
Confidence Assessment:
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Given Solution:
`a** Position at clock time is x = Acos(`omega* t)
Velocity = -`omega *A*sin(`omega* t)
Accel = -`omega * A * cos(`omega* t)
University Physics students should note that velocity and acceleration are the first and second derivatives of the position function. **
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Self-critique (if necessary):
thats at equilibriu m if i remember correctly.
Self-critique Rating:
Question: **** `qHow is the acceleration of the pendulum related to the centripetal acceleration of the point on the reference circle?
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Your solution:
a = v^2/r and so its wr or wa
Confidence Assessment:
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Given Solution:
`aSTUDENT ANSWER: a = -`omega A sin(`omega *t) and
aCent = v^2/r for the circle modeling SHM
INSTRUCTOR AMPLIFICATION:
** The centripetal acceleration of the point on the reference circle, which acts toward the center of the circle, has two components, one in the x direction and one in the y direction. The component of the centripetal acceleration in the direction of the motion of the oscillator is equal to the acceleration of the oscillator.
If the oscillator is at position theta then the centripetal acceleration has direction -theta (back toward the center of the circle, opposite to the position vector). The centripetal acceleration is aCent = v^2 / r; so the x and y components are respectively
ax = aCent * cos(-theta) = v^2 / r * cos(theta) and
ay = aCent * sin(-theta) = -v^2 / r * sin(theta). **
How is the kinetic energy of the pendulum related to its restoring force constant k, the amplitude of its motion, and its position x?
** The PE of the pendulum at displacement x is .5 k x^2.
By conservation of energy, if nonconservative forces are negligible, we find that the KE of the pendulum at position x is.5 k A^2 - .5 k x^2. This result is obtained from the fact that at max displacement A the KE is zero, and the KE change from displacement A to displacement x is the negative of the PE change between these points.
Thus .5 m v^2 = .5 k A^2 - .5 k x^2. Solving for v we have
v = +- sqrt( .5 k / m * (A^2 - x^2) ) . **
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Self-critique (if necessary):
oh yeah , find the x and y components of acceleration. usually looking for the x component to relate refference circle to a pendulum
i have these relationships written down from the book. got them explained from class notes, i shold have it set as knowledge , the reference circle wasnt really explained in book bit i see how it is used .
Self-critique Rating:
Question: **** `qHow can we determine the maximum velocity of a pendulum using a washer and a rigid barrier?
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Your solution:
rigid barrier makes me think of physical pendulum ,
without the details of the length of the pendulum and the displacement we cannot find the max velocity
i dont know how the washer is involved with this
Confidence Assessment:
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Given Solution:
`aGOOD STUDENT ANSWER: If we pullback a pendulum of length L a distance x (much smaller than L), and stop the motion at the equilibrium point (vertical limit of motion) a washer on the pendulum will become a projectile and project off the pendulum, to land at a distance from which we can determine the horizontal velocity of the washer. That velocity is the same as the max velocity of the pendulum, since the max velocity is that which is at the lowest point in its path.
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Self-critique (if necessary):
yeah the projectile motion of the washer will tell us the horizontal velocity and this is the vmax
Self-critique Rating:
Question: **** `qPrinciples of Physics and General College Physics Problem 11.3. Springs compress 5.0 mm when 68 kg driver gets in; frequency of vibration of 1500-kg car?
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Your solution:
k=-68kg*9.8m/s^2/.005= 133280N/m
the frequency is found , 1/2pi*sqrt(133280N/1568kg)= 1.47cycles /second
Confidence Assessment:
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Given Solution:
`aFrom the weight of the driver and the compression of the spring, we determine the spring constant (the 'stiffness' of the spring in N / m):
driver weight of 68 kg * 9.8 m/s^2 = 670 N compresses the spring .05 meters, so since | F | = k | x | we have k = | F | / | x | = 670 N / (.005 m) = 134 000 N / m.
Now from the force constant and the mass of the system we have
omega = sqrt(k / m) = sqrt( (134 000 N/m) / (1570 kg) ) = 9 sqrt( (N/m) / kg) ) = 9 sqrt( (kg / s^2) / kg) = 9 rad / s, approximately, or about 1.5 cycles / second.
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Self-critique (if necessary):
different order but pretty much the same thing
Self-critique Rating:
Question: **** `qPrinciples of Physics and General College Physics problem 11.30: Pendulum with period 0.80 s on Earth; period on Mars, where acceleration of gravity is 0.37 times as great.
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Your solution:
the period .8s= sqrt(9.8/L) gives L= 9.8/.8^2= 15.3 m
the accelratio of gravity on mars is .37*9.8 = 3.63m.s^2
the perid on mars is sqrt(3.63/15.3)= .469s
Confidence Assessment:
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Given Solution:
`aThe period of a angular frequency harmonic oscillator is sqrt(k / m), and the time required for a cycle, i.e., the period of the cycle, is the time required to complete a cycle of 2 pi radians.
For a pendulum we have k = sqrt( m g / L ), where g is the acceleration of gravity. Thus for a pendulum omega = sqrt(k / m) = sqrt( (m g / L) / m) = sqrt( g / L).
From this we see that for a given length, the frequency of the pendulum is proportional to sqrt(g). The period is inversely proportional to the frequency, so the period is inversely proportional to sqrt(g).
Thus we have
period on Mars / period on Earth = sqrt( gravitational acceleration on Earth / gravitational acceleration on Mars) = sqrt( 1 / .37) = 1.7, approximately. So the period on Mars would be about 1.7 * .80 sec = 1.3 sec, approx.
As an alternative to the reasoning or proportionality, we can actually determine the length of the pendulum, and use this length with the actual acceleration of gravity on Mars.
We have
period = 2 pi rad / angular frequency = 2 pi rad / (sqrt( g / L) ) = 2 pi rad * sqrt(L / g). We know the period and acceleration of gravity on Earth, so we can solve for the length:
Starting with period = 2 pi sqrt(L / g)) we square both sides to get
period^2 = 4 pi^2 L / g. Multiplying both sides by g / (4 pi^2) we get
L = g * period^2 / (4 pi^2) = 9.8 m/s^2 * (0.80 sec)^2 / (4 pi^2) = .15 meters.
The pendulum is .15 meters, or 15 cm, long.
On Mars the acceleration of gravity is about 0.37 * 9.8 m/s^2 = 3.6 m/s^2, approx.. The period of a pendulum on Mars would therefore be
period = 2 pi sqrt(L / g) = 2 pi sqrt(.15 m / (3.6 m/s^2)) = 1.3 seconds, approx.
This agrees with the 1.3 second result from the proportionality argument.
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Self-critique (if necessary):
the direct reasonoing makes this much more simple. i guess it makes sense that if the proportion of mars accel is .37 times earth , than the period sqrt g/L) would share the same proportionality.
The proportionality of periods would be 1 / sqrt(.37).
Be sure you understand why.
Self-critique Rating:
Question: **** `qQuery gen problem 11.14 80 N to compress popgun spring .2 m with .15 kg ball.
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Your solution:
the k constant is (80N+9,8*.15kg)/.2m = 407 N/M
the angular frequency is found to be sqrt(407N/8.31kg) = 7rad/s
Confidence Assessment:
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Given Solution:
`a** The PE of the system will be .5 k A^2, where A = .2 m and k = F / x = 80 N / (.2 m) = 400 N / m.
The KE of the released ball will in the ideal case, which is assumed here, be .5 m v^2 = .5 k A^2. Solving for v we obtain
v = +- sqrt( k A^2 / m ) = +- sqrt( 400 N/m * (.2 m)^2 / (.15 kg) ) = +- sqrt( 106 m^2 / s^2) = +-10.3 m/s, approx.
The speed of the ball is the magnitude 10.3 m/s of the velocity. **
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Self-critique (if necessary):
i figured the 80 N force was added to the balls gravitational force on the spring.assuming vertical setup,
needed to find the speed of the ball when released. . this is founf from ke and pe relationshi to get v
this was in the book and now i understand why i got it wrong. i didint know why,
Self-critique Rating:
Question: **** `qQuery gen phy problem 11.24 spring 305 N/m amplitude 28 cm suspended mass .260 kg.
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Your solution:
the angular frequency is sqrt(305/.26)= 34.3rad/s
the maximum velocity will occur at wA = 34.3 rad.s* .28m= 9.59 m/s
Confidence Assessment:
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Given Solution:
`a**The solution given here is for restoring force constant 210 N/m and mass .250 kg. You should be able to adapt your solution accordingly, and you should understand why the angular frequency will be sqrt(305 * .250 / (210 * .260)) times as great as that given here.The angular frequency of the oscillation (the angular velocity of the point on the reference circle) is
omega = sqrt(k / m),
with k = 210 N/m and m = .250 kg.
The equation of motion could be y = A sin(omega * t).
We obtain omega = sqrt( 210 N/m / (.250 kg) ) = sqrt( 840 s^-2) = 29 rad/s, approx..
A is the amplitude 28 cm of motion.
So the equation could be
y = 28 cm sin(29 rad/s * t).
The motion could also be modeled by the function 28 sin (29 rad/s * t + theta0) for any theta0. The same expression with cosine instead of sine would be equally valid, though for any given situation theta0 will be different for the cosine model than for the sine model. **
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Self-critique (if necessary):
mine works for only maximum velocit , if we want the velocity function we need to find the x or y component of position.
you found the y component with the starting angle with respect to this.
Self-critique Rating:
Question: **** `qUniv. 13.74 (13.62 10th edition). 40 N force stretches spring .25 m. What is mass if period of oscillation 1.00 sec? Amplitude .05 m, position and vel .35 sec after passing equil going downward?
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Your solution:
the k constant is 40N / .25m = 160 n/m
T=1s= 2pi*sqrt(160/m) so the mass= 160/(1/2pi)^2=6316.55kg
w = sqrt(160/6316.55)= .159rad/s
the position is found to be .05*cos(.159*.35)= .05radians below equilibrium
the velocity is -.159*.05*sin(.159*.35)= .00000662 m/s
doubting this slowness but it could be true
Confidence Assessment:
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Given Solution:
`aGOOD PARTIAL STUDENT SOLUTION WITH INSTRUCTOR COMMENT
I am sorry to say I did not get that one--but mostly because I am hurrying through these, and I could not locate in my notes, altough I remember doing extensive work through the T=period problems--let me look at Set 9 for a moment.
I think I found something now. If `omega = 2`pi/T and t = 1 sec, `omega = 2pi rad/s
If I convert to accel, thenI can find the mass by way of F = ma.
a = `omega ^2 * A. I do not know A yet so that is no good.
}If A = x then my pullback of x = .25 m would qualify as A, so
a = (2`pi rad/s) ^2 * .25 m = 9.87 m/s^2
So m = F/a = 40.0N/9.87 m/s^2 = 4.05 kg
THAT IS PART A.
INSTRUCTOR COMMENT:
** Good. But note also that you could have found m = k / omega^2 from omega = sqrt(k/m).
F = -k x so 40 N = k * .25 m and k = 160 N/m.
Thus m = 160 N/m / (2 pi rad/s)^2 = 4 kg approx..
STUDENT SOLUTION TO PART B:For part B If A = .050m and T = 1 sec, then the position can be found using the equation, x = A cos(`omega *t)
INSTRUCTOR COMMENT:
** You could model this situation with negative omega, using x = .05 m * sin(-omega * t). This would have the mass passing thru equilibrium at t = 0 and moving downward at that instant.
Then at t = .35 s you would have x = .05 m * sin( - 2 pi rad/s * .35 s ) = .05 m * sin( -.22 rad) = -.040 m, approx..
Velocity would be dx/dt = - 2 pi rad/s * .05 m * cos(-2 pi rad/s * .35 s) = -.18 m/s, approx..
Alternatively you might use the cosine function with an initial angle theta0 chosen to fulfill the given initial conditions:
x = .05 m * cos(2 pi rad/s * t + theta0), with theta0 chosen so that at t = 0 velocity dx/dt is negative and position is x = 0.
Since cos(pi/2) and cos(3 pi/2) are both zero, theta0 will be either pi/2 or 3 pi/2.
The velocity function will be v = dx/dt = -2 pi rad/s * .05 m sin(2 pi rad/s * t + theta0). At t = 0, theta0 = pi/2 will result in negative v and theta0 = 3 pi/2 in positive v so we conclude that theta0 must be pi/2.
Our function is therefore
x(t) = .05 m * cos(2 pi rad/s * t + pi/2).
This could also be written
x(t) = .05 m * cos( 2 pi rad/s * ( t + 1/4 sec) ), indicating a 'time shift' of -1/4 sec with respect to the function x(t) = .05 m cos(2 pi rad/s * t). **
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Self-critique (if necessary):
i had a problem with choosing the theta 0 up or down ar pi/2 and 3pi/2 so they work. they are zero so i thought that adding them would be pointless and left them out. but they are needed as a starting point for w't to relate to)
should have found omega to be equall to 2pirad/s
that was a no brainer too
Self-critique Rating:
Question: **** `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.
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Self-critique (if necessary):
Self-critique Rating:
Good work and good self-critiques. See my note(s).