ch. 9_1 quiz
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ch. 9_1 quiz
1) Uniform circular motion in the xy plane is projected onto the xaxis 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 yaxis would (not) result in SHM, however with different initial (t = 0) condition (see sheet 4').
2) Uniform circular motion in the xy plane is projected onto the xaxis 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!
2) Uniform circular motion in the xy plane is projected onto the xaxis 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!
guest_01 Guest
Re: ch. 9_1 quiz
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
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
guest1 Guest
Re: ch. 9_1 quiz
I tried using your solution to question 4 and it marked it wrong. anyone know another way to do it?
guesttt Guest
ques 3,4
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?
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?
kath Guest
quest 4
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
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
confused Guest
Here's what I got....
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...
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...
An Guest
question 3,4
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
kacy Guest
ques 3&4
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..
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..
confused Guest
Re: ch. 9_1 quiz
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.
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.
Guest01 Posts : 133
Join date : 20080919
Re: ch. 9_1 quiz
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
(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
heyy Guest
Re: ch. 9_1 quiz
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.
The time is the decimal, in seconds.
Guest01 Posts : 133
Join date : 20080919
Question regarding question 3
For question number 4 why do we disregard the time associated with the velocity?
Thanks
Thanks
BGGirl Posts : 6
Join date : 20080921
why does #4 not work?
#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?
wtf Guest
Q4
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
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
cyn Guest
Re: ch. 9_1 quiz
switch your calc mode to radians for #3 and 4 and use the equations in the post with the red
an Guest
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