# Ch 9.2 Help

## Ch 9.2 Help

Question 1: (1 point)

An object in SHM with a period of 7.741 s has a maximum velocity of 4.957 m/s. What is the amplitude (see sheet 6)?

(How to solve):

Question 2: (1 point)

An object in SHM has a maximum velocity of 11.86 m/s and an amplitude of 0.6553 m. What is the magnitude of the maximum acceleration (see sheet 6)? Indicate with a positive (negative) sign whether the acceleration as a function of time can (cannot) be written as -(2 /T)2x, where x is the displacement as a function of time and T is the period.

(How to solve):

Question 3: (1 point)

A spring is compressed by 10.42 cm when a 13.644 N force is exerted. What is the spring constant (see sheet 14)? Indicate with a positive (negative) sign whether the spring is stretched by the same (a different) amount when a 13.644 N tension acts on it.

(How to solve):

Question 4: (1 point)

A block attached to a spring with a spring constant of 40.41 N/m oscillates frictionless with a frequency of 3.898 Hz on a horizontal plane. What is the mass of the block (see sheet 15)?

(How to solve):

sedwards
Guest

1.+
2.+
3.+
4.+

freshfig
Guest

## How to solve

1. Use eqn: V0= (2pi/T)*X0
X0=amplitude
V0=max. velocity
T = Period
X0=(V0*T)/(2pi)

2. Need to use two eqns for this one. Use the V0=(2pi/T)*X0 to find period.
T= (2*pi*X0)/(V0)
Use whatever you get for T and plug into the A0 eqn:
A0=((2pi/T)^2)*X0

3. F=-kX
k=F/x

4. Use:
Remember the frequency and period are related to each other.
T=(1/f)
T= 2pi* sqrt(m/k)
solve for m
m= ((T/2*pi)^2)*k

Hopes this helps!!!

freshfig
Guest

## Re: Ch 9.2 Help

for question four What is m and what is k?

Posts : 22
Join date : 2008-09-17

## Re: Ch 9.2 Help

You are looking at two different formulas for Question 4.

We are looking for m, k is the spring constant.

Guest01

Posts : 133
Join date : 2008-09-19

## Re: Ch 9.2 Help

Can you show how you solved the questions posted so I can see how you are getting the ans??

Meg
Guest

## Re: Ch 9.2 Help

Okay, let's just completely break these two questions down. There are probably two reasons why you are getting these wrong. The first is that your calculator is programmed to calculate angles in degrees. We do NOT want that. We want it to be in Radians. If you are unsure of this, or do not know how to change it, for TI-83 Plus calculators hit the Mode button. It is right next to the yellow 2nd button. A list will come up and the third choice down should be Radians Degrees. Make sure Radians is highlighted by having a black box over it. That is all.

If you do not know why it is in Radians, or the difference between Radians and Degrees, or what exactly is a Radian, just go ahead and ask.

I do assume that anyone reading this does know, however I am purposely being redundant, not to be an ass but just in case someone accidentally overpasses this information or forgets it.

The second reason you are probably getting a wrong answer is you are not treating (that's 2*pi) by itself, i.e. is acted as a constant here. Just like χ represents some type of variable so does π. It may seem obvious but if you are doing the calculations improperly by missing out on certain things. Basically what is going on is 2 * π/q or a bit clearer 2 * (π/q); where q just represents some type of number. It should be calculated this way (2 * π) / q i.e. calculate first then divide into/by it. Sometimes formulas will be different such as or whatever, but it should be treated as one variable, in loose terms.

It may be note worthy to say multiplication is cumulative, meaning now matter which numbers you multiply with first, it will all equal the same thing. For example: 2π*3 can calculated as 2*π*3, π*2*3, 3*2*π, 3*π*2, π*3*2, 3*π*2 and it should all equal the same thing. You should get a number like 18.8496.

Try these two examples:
(1) 3π/8 (2) 4π5
1. (3π)/8
1. =
Spoiler:
1.1781

2. 4*π*5
2. =
Spoiler:
62.8319

Armed with this information let's try our problems. Here were my submitted calculations.

Question 2
An object in SHM has a maximum velocity of 25.61 m/s and an amplitude of 0.5737 m. What is the magnitude of the maximum acceleration (see sheet 6)? Indicate with a positive (negative) sign whether the acceleration as a function of time can (cannot) be written as -(2π/T)2x, where x is the displacement as a function of time and T is the period.
Spoiler:
1143.2318

To Solve:
We know the true and false question is Positive.

The question is asking for the Maximum Acceleration, so from our formulas which one can we substitute with the numbers given? We can use the formula A0=((2π/T)2)*X0.

"Wait just a darn minute there! We don't have our period!" you say?
Ah, we don't, but we can calculate it from the information given. Remember the formula to find the Period? Well here it is T= (2*π*X0)/(V0), we get by rearranging the formula V0=(2π/T)*X0 Now substitute those numbers. Let's go back and find out what is what.

Here's what we have/what we need:
And for fun, let's COLOR Code them.
X0 = 0.5737
V0 = 25.61 m/s
=
T = ?
A0 = ?

Okay now the calculations:
T = (*X0)/V0
T = (*0.5737)/25.61 m/s
T = (3.6047/25.61 m/s
T = 0.1408

Now:
X0 = 0.5737
V0 = 25.61 m/s
=
T = 0.1408
A0 = ?

So use the first formula to get the answer we are looking for:
A0 = ((/T)2)*X0
A0 = ((/0.1408)2)*0.5737
A0 = ((44.6249)2)*0.5737
A0 = (1991.3817)*0.5737
A0 =
Spoiler:
1142.4557

The answers are slightly different most likely due to rounding. I suggest using as many numbers as possible.

Try that for now. See if you can get Question 4 before I can even post it

Guest01

Posts : 133
Join date : 2008-09-19

## Re: Ch 9.2 Help

For these problems, it doesn't matter what mode your calculator is in, since we are not working with degrees or radians. If you are still having trouble, it probably has to do with the 2*pi, make sure you write it like (2*pi).

J
Guest

i got #4 wrong

hhh
Guest

## question 4

To do question 4:

m= T^2*k / 4pi^2

zero
Guest

## number 4???

#4 is still not working for me

physicss
Guest