Question

This is for a Harmonic Motion function. (Pendulum)My last tutor helped me to the very finish line and ended the session and I miss clicked and couldn’t see the answers.... so can you help me finish the rest? (If we get through this entire section: I will write a very long letter of gratitude and give you the highest rating for helping me understand. I truly just want to finish) There are only short questions left, which are mainly just SENTENCES... I know we can do this. Here is some background information of what I got so far :•Period of pendulum:1.6•B value (rounded to the nearest thousandth): 3.762 rad/s•Amplitude, in meters(rounded to the nearest thousandth):0.733m•Trigonometric function that starts at its peak of amplitude: Sine•Harmonic motion formula for the base scenario: x(t)=0.733sin(3.762t)•statement for the zeros of the base harmonic motion function: “When the net force acting on the pendulum is zero, the base harmonic motion will cease. The pendulum will be at its equilibrium if the force is nothing. This means that, at the equilibrium position restoring force due to gravity in the cord, the pendulum becomes identical. (At time (t=0), the position of the equilibrium is at (x=0), which is the equilibrium position of the pendulum. •where the pendulum is located at 10 seconds. Including the resting position, rounded to the nearest thousandth: “substituting the values in the equation X(10)=0.733sin(3.762x10)X(10)=0.733(-0.079)X(10)=-0.058mSo if the starting angle of the pendulum is Ø=60 degrees and Ø=pi/3, the initial function of the pendulum would be x(t)=0.733sin(3.762t+ pi/3) “ WHEN I CHANGED THE STARTING ANGLE The tutor helped me get the function that represents the position of the pendulum eh it started from thirty which is : •the harmonic motion function that represents the position of the pendulum when it started from 30 degrees: x(t)=0.733sin(3.726t + pi/6. WHAT I NEED BEFORE MOVING FORWARD: A prediction on how that new starting angle of 30 degrees affected the pendulum and its harmonic function? We need this in full sentences. (Since we already know that the harmonic motion function that represents the position of the pendulum when it started from 30 degrees is x(t)=0.733sin(3.726t + pi/6 ) ( we will need a comparison to THE BASE SCENARIO of 60 degrees . Was your prediction correct? Please explain in full sentences. ) STEP THREE: Time to change the length of the pendulum (the starting angle will be 60 degrees, but the length of the pendulum will be shortened to 0.4 meters.)How do you think shortening the pendulum to 0.4 m will affect the pendulum and it's equation? Explain in a few sentences what is the harmonic motion function for the pendulum when the length of the pendulum is shortened to 0.4 meters? (I GOT A LITTLE BIT OF THIS : L=0.4m, length of pendulum T=2pi sqrtL/g, period of pendulum T= 2 x 3.14 Please help finish the rest. (This is all I got from the previous tutor before the session ended and I miss clicked the viewing of answers) Was your prediction correct? explain how changing the length of the pendulum affected the function. Why did changing the length of the pendulum affect the function the way that it did. We need to answer this in a full sentences. THE HOMESTRETCH STEP 4 : changing the period (start with your base harmonic motion function from the first time. Keeping the same a value, replace the b value with 4.5)explain how replacing the b value in the base harmonic motion function from stage first time , will affect it. Answer in a few sentences. What is the amplitude of the new harmonic motion function. Explain in full sentences. what is the period of the new harmonic motion function? Explain. Using the new harmonic motion function, how many periods will the pendulum complete over the course of 60 seconds? Explain. (If you made it this far, I would like to thank you for your service and patience with my tedious questions. Thank you for granting me the ability to see the understanding of these questions. You are greatly appreciated)

Answer 1

Given data:

Function of simple harmonic motion is,

`x(t)=A\sin (\omega t+\phi)`

Here,

`\phi\text{ is initial angle}`

1. Function given for simple pndulum is,

`x(t)=0.733\sin \text{ (3.736t+}(\Pi)/(6)\text{)}\ldots(1)`

If the initial or starting angle of pendulum is larger, restoring force acting on pendulum is larger so, that pedulum travels more faster than those pendulum which have small starting angle.

Position function of pendulum for starting angle is as equation-(1).

2. Pendulum which have staring angle is 60 degrees which have moves more faster as compare to pendulum which have starting angle is 30 degrees. Period of pendulum does not affected by starting angle. But, Amplitude of pendulum is larger for larger strting angle, means that amplitude of pendulum is larger when starting angle is 60 degrees.

Step-3:

1.Given length of pendulum,

`l=0.4\text{ m}`

Formula of period of pendulum,

`T=2\Pi\sqrt[]{(l)/(g)}`

If length of pendulum is shortening, according to above equation period of pendulum will decreases hence, pendulum will moves slower so, that angular frequency inceases.

2.Pendulum will moves faster as if length of pendulum decreases.

now, substitute value of length in above equation,

`\begin{gathered} T=2*3.14*\sqrt[]{\frac{0.4\text{ m}}{9.8m/s^2}} \\ T=1.27\text{ seconds} \end{gathered}`

New, angular frequency of pedulum is as follows:

`\begin{gathered} \omega=(2\Pi)/(T) \\ \omega=\frac{2*3.14}{1.27\text{ s}} \\ \omega=4.944\text{ rad/s} \end{gathered}`

Now, new position function of pendulum is,

`x(t)=0.733\sin (4.944t+(\Pi)/(6))`

3. due to change in length of pendulum, angular velocity of pendulum changes beculase it has direct relation as below,

`\omega=\sqrt[]{(g)/(l)}`

Hence, if angular frequency of pedulum changes, its function of position is also changes.

Step:4

1. Given function for pendulum is,

`x(t)=0.733\sin \text{ (3.762t)}`

Changing the values of b means angular frequency,

`\omega=4.5\text{ rad/s}`

In the given case, angular frequency of pendulum increases, means that velocity of pendulum will increases.

formula:

`v=\omega\sqrt[]{A^2-x^2}`

Acceleration of pendulum also increases, due to increases in angular velocity.

formula:

`a=-A\omega^2\sin (\omega t)`

2. Amplitude of harmonic motion will not depends on the angular frequency of motion. Hence, it remains constant. Amplitude of new harmonic function is as below,

`A=0.733\text{ m}`

3. Priod of pendulum will decreases, if angular freqnecy of decreases. Hence, pendulum moving faster.

period of new harmonic function is as follows:

`\begin{gathered} T=(2\Pi)/(\omega) \\ T=(2*3.14)/(4.5) \\ T=1.395\text{ seconds} \end{gathered}`

4. Number of periods compeleted by pendulum will increases due to increases in angular frequency of pendulum beacuse of it period of pendulum will decreases.

Formula:

`n=(t)/(T)`

Number of periods compeleted by pendulum is 60 s is as follows:

`t=60\text{ seconds}`

Hence,

`\begin{gathered} n=\frac{60\text{ s}}{1.395\text{ }s} \\ n=43.01 \\ n\cong43 \end{gathered}`

Number of periods compeleted by the pendulum is 43.