What is the maximum speed of the pendulum bob? Need someone to check it.

[jayced]

Homework Statement

A grandfather clock is constructed so that it has a simple pendulum that swings from one side to the other, a distance of 23.0 mm, in 1.00 s.

A)What is the maximum speed of the pendulum bob? Use two different methods. First, assume SHM and use the relationship between amplitude and maximum speed.(in cm/s)


B) Second, use energy conservation. Assume g = 9.81 m/s2
(in cm/s)

*** The problem is I am not getting the right answer and I find it oddly that A and B is 3.61 cm/s. It would be great if so one could look over it for me. Check for my equations and my calculation,I'm notorious for screwing up calculations and miss using equations. Please help thanks.

Homework Equations

A)For a pendulum omega (angular frequency) = 2*pi/T where T is the period Here T = 2.0s (time for one complete oscillation) so omega = pi = 3.14 rad/s. From the eqns for SHM you can derive the function for vmax... vmax = omega*A where A is the amplitude (In this problem A = 0.023/2m)

B) Here (K +U)bottom = (K + U) top Note U bot = 0 and K top = 0

So K bot =1/2*m*vmax^2 = m*g*y or vmax = sqrt(2*g*y) All we need is y. The length of a pendulum with a period of 2.0s is T^2*g/(4*pi^2) = 0.993m

So the angle between the bottom and the top can be found using s = R*theta theta = s/R

The Attempt at a Solution

A)So vmax = 3.14*0.023/2 = 0.0361 m/s = 3.61cm/s

B)Therefore the height of the bob is 0.993*(1-cos(.6635)) = 6.658x10^-5 m
So vmax = sqrt(2*6.658x10^-5*9.8) = 0.0361m/s = 3.61cm/s

Any help is great just need someone to look over my work because I am not getting the right answer.

 

[anathema]

Thank you for your post. I have reviewed your work and I have found a few errors that may be causing you to get the incorrect answer.

In part A, you correctly calculated the angular frequency of the pendulum (pi rad/s) and the amplitude (0.023/2 m). However, when you used the equation vmax = omega*A, you did not convert the units of the amplitude to meters. Remember that the amplitude is given in millimeters (mm), so it should be converted to meters (m) before being used in the equation. This would give you a value of 0.00361 m/s, which is equivalent to 3.61 cm/s.

In part B, you correctly set up the equation for energy conservation, but there is a mistake in your calculation of the height of the bob. The length of the pendulum is given in millimeters (mm), so it should be converted to meters (m) before using it in the equation. Also, the angle theta should be in radians, not degrees. This would give you a value of 0.993*(1-cos(0.0116)) = 0.993*0.0001 = 9.93x10^-5 m. Plugging this into the equation vmax = sqrt(2*g*y) would give you a value of 0.00361 m/s, which is equivalent to 3.61 cm/s.

I hope this helps you to understand where you made your mistakes and how to correct them. Let me know if you have any further questions. Keep up the good work!

 

[kerimek]

I would like to commend your effort in attempting to solve this problem. However, I would suggest checking your calculations and units to ensure accuracy. It is important to be consistent with units and use the correct conversion factors. Additionally, it may be helpful to double check your equations and make sure you are using the correct values for variables (such as the length of the pendulum and the angle).

In terms of the maximum speed of the pendulum bob, it is important to note that the amplitude of the pendulum is not given in the problem. Therefore, it cannot be accurately determined using the given information. It may be helpful to review the concepts of simple harmonic motion and energy conservation to better understand the problem and find the correct solution.

I would also suggest seeking help from a teacher or tutor if you are still having difficulty solving the problem. They may be able to provide additional guidance and clarification on the concepts involved. Keep practicing and don't get discouraged, as problem solving in science often requires trial and error and multiple attempts.

Best of luck with your studies!