ch. 9_1 quiz

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ch. 9_1 quiz

Post  guest_01 on Sat Oct 25, 2008 10:26 pm

1) Uniform circular motion in the x-y plane is projected onto the x-axis resulting in an oscillation, simple harmonic motion (SHM). If the radius of the circle is 0.1535 m what is the amplitude of the oscillation (see sheet 3)? Indicate with a negative (positive) sign whether the projection onto the y-axis would (not) result in SHM, however with different initial (t = 0) condition (see sheet 4').

2) Uniform circular motion in the x-y plane is projected onto the x-axis resulting in an oscillation, simple harmonic motion (SHM). If the constant angular velocity of the circular motion is 3.645 rads/s what is the frequency of the oscillation (see sheet 4)? Indicate with a negative (positive) sign whether the period can (cannot) be calculated too from the data given.

3) An object in SHM with a period of 5.465 s is at 0.7783 s a distance of 0.977 m displaced from the equilibrium position (displacement = 0). What is the amplitude of the oscillation (see sheet 4)? Indicate with a positive (negative) sign whether at t = 0 the displacement is minimal (maximal) Note! in x = x0cos(wt), always use the full argument (wt), including the t. The quantity in the (...) is an angle in radians (watch your calculator!).

4) An object in SHM with a frequency of 1.918 Hz (see sheet 9) has a velocity of (-2.931) m/s at 0.1373 s. What is the magnitude of the maximum velocity (see sheet 6)?

Solutions please? Thanks!

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Re: ch. 9_1 quiz

Post  guest1 on Sun Oct 26, 2008 2:33 am

Questions 1, 2 and 3 are negative.

Q1: the answer is equal to the radius

Q2: use v/(2*pi)

Q3: x=Acos(2*pi*f*t)

Q4: There's probably an easier way for this one, but this is what I did and it was right:

v_max= A(sqrt(k/m))

to find sqrt(k/m) : f=(1/(2*pi))*(sqrt(k/m)) , so sqrt(k/m)=f*(2*pi)

to find A: v_x=-2*pi*f*A*sin(2*pi*f*t)
solve for A, then plug that into this equation: v_max=A*2*pi*f

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Re: ch. 9_1 quiz

Post  guesttt on Sun Oct 26, 2008 9:56 am

I tried using your solution to question 4 and it marked it wrong. anyone know another way to do it?

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Re: ch. 9_1 quiz

Post  Guest 00 on Sun Oct 26, 2008 12:33 pm

Just use VmAx = V / sin(2pi*f*t) make the given velocity positive

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ques 3,4

Post  kath on Sun Oct 26, 2008 12:44 pm

I got both 3 and 4 wrong, can you tell me what I did incorrect?

3.) An object in SHM with a period of 7.481 s is at 0.4188 s a distance of 1.31 m displaced from the equilibrium position (displacement = 0). What is the amplitude of the oscillation (see sheet 4)? Indicate with a positive (negative) sign whether at t = 0 the displacement is minimal (maximal) Note! in x = x0cos(wt), always use the full argument (wt), including the t. The quantity in the (...) is an angle in radians (watch your calculator!).

x=Acos(2pi*f*t)
1.31 = Acos(2pi*7.481*.4188)
A=1.95
(And then I put it as negative)

4.) An object in SHM with a frequency of 2.692 Hz (see sheet 9) has a velocity of (-4.911) m/s at 0.0858 s. What is the magnitude of the maximum velocity (see sheet 6)?

sqrt (k/m) = f*2pi = 2.692 * 2pi = 16.9143
vx = -2pi*f*A*sin(2pi*f*t)
-4.911 = -2pi * 2.692 * A * sin(2pi * 2.692 * .0858)
A = 11.464
v_max = 11.464 * 16.91 = 193.9
and that was put as positive

what did i do wrong?

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quest 4

Post  confused on Sun Oct 26, 2008 12:57 pm

if we do question 4 with the formula v/sin(2pi*f*t), dont we come out with a negative answer?

An object in SHM with a frequency of 2.699 Hz (see sheet 9) has a velocity of (-3.707) m/s at 0.1016 s. What is the magnitude of the maximum velocity

(-3.707)/sin(2pi*.1016*2.699)
= -123.29

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Here's what I got....

Post  An on Sun Oct 26, 2008 1:54 pm

Signs

1) -
2) -
3) -
4) +

1) Just the radius
2) V/2(pi)
3) x= A cos [(2(pi)t)/PERIOD]
4) V/SIN (2(pi)Ft)


good Luck...

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Re: ch. 9_1 quiz

Post  hello on Sun Oct 26, 2008 3:05 pm

i keep getting # 4 wrong

do we have to convert hertz of something?

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question 3,4

Post  kacy on Sun Oct 26, 2008 3:52 pm

I used the formulas posted and they were correct i think you just need to change the mode of your calculator to radians like question number 3 said too and you will get them right. hope that helps

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ques 3&4

Post  confused on Sun Oct 26, 2008 8:56 pm

I've done the quiz like 5 times already and I keep getting 3 and four wrong, even though I'm using the equations given and switching my calculators to the different modes...can someone please help me?

ques 3 An object in SHM with a period of 6.131 s is at 0.4277 s a distance of 0.616 m displaced from the equilibrium position (displacement = 0). What is the amplitude of the oscillation (see sheet 4)? Indicate with a positive (negative) sign whether at t = 0 the displacement is minimal (maximal) Note! in x = x0cos(wt), always use the full argument (wt), including the t.

.616 = Acos (2pi*.4277/6.131)
A= .61602


ques 4= An object in SHM with a frequency of 1.831 Hz (see sheet 9) has a velocity of (-2.402) m/s at 0.1455 s. What is the magnitude of the maximum velocity

(2.402)/sin( 2pi*1.831*.1455)
= 2.4148

are you guys getting different answers then these? I really don't know what I'm doing wrong..

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Re: ch. 9_1 quiz

Post  Guest01 on Sun Oct 26, 2008 9:15 pm

You have gone awry in your calculations, I am getting different answers. For Question 3, first find (2*pi*t) then divide that by the period. Then find its cosine and divide it into the distance.

For Question 4 they give you frequency but you need the period. Period is equal to (1/f) where f is the frequency. make sure you use this as a fraction in your calculation and not mess it up as you did in the previous question. Then find its sine and divide it into the velocity. Even though it is given as negative, you can ignore the minus and treat it as positive for this question.

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Re: ch. 9_1 quiz

Post  guest_ on Sun Oct 26, 2008 9:33 pm

as mentioned above, your calculator needs to be in RADIANS

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Re: ch. 9_1 quiz

Post  heyy on Sun Oct 26, 2008 9:51 pm

so can u guys substitute hte #'s to make it easier to us

(2*pi*t) then divide that by the period. Then find its cosine and divide it into the distance.

so t is what, 5.465 or .7783
and which one is period?

5.465 s is at 0.7783 s a distance of 0.977 m

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Re: ch. 9_1 quiz

Post  Guest01 on Sun Oct 26, 2008 10:03 pm

It says the SHS (simple harmonic motion) occurs at the period/frequency of 5.xxx s (paraphrasing here). There is the period.
The time is the decimal, in seconds.

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Re: ch. 9_1 quiz

Post  mmm on Sun Oct 26, 2008 10:30 pm

how u do 3

helpppp Mad

mmm
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Question regarding question 3

Post  BGGirl on Sun Oct 26, 2008 10:43 pm

For question number 4 why do we disregard the time associated with the velocity? scratch


Thanks

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Re: ch. 9_1 quiz

Post  aa on Sun Oct 26, 2008 11:09 pm

plz help with 3

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why does #4 not work?

Post  wtf on Sun Oct 26, 2008 11:53 pm

#4 keeps coming out wrong for me i've done it like 6 times and i've done it exactly how everyone has said to do it so does anyone else have any ideas? scratch

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Q4

Post  cyn on Mon Oct 27, 2008 12:36 am

Ok they ask for
V= ? ( magnitud of velocity)
so:
V= -(V_o)(sin(2pi*f*t))
ex:
given:
f= 2.499 HZ
V(V_o)= -3.747m/s (V_o is always linear m/s)
t=.1062 s
so
V= -(-3.747m/s)(sin(2pi*2.499*.1062))
V=3.729
Ps. remember there is more than one way to use w, 3:
w=2(pi)*f
w= 2(pi)/ T
w= theta/t
Hope this helps you

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questions 3 & 4

Post  guest1 on Mon Oct 27, 2008 12:52 am

you have to put ur calculator in radian mode for both questions

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Q4

Post  cyn on Mon Oct 27, 2008 12:52 am

V_o= max veloc.
V= magnitud of maximum velocity.

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Re: ch. 9_1 quiz

Post  an on Mon Oct 27, 2008 1:31 am

switch your calc mode to radians for #3 and 4 and use the equations in the post with the red

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Re: ch. 9_1 quiz

Post  Guest01 on Mon Oct 27, 2008 2:27 am

Have you checked above?

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