# ch. 9_1 quiz

## ch. 9_1 quiz

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!

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!

**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

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 : 2008-09-19

## 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 : 2008-09-19

## 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 : 2008-09-21

## 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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