An object moves up and down the y-axis with an acceleration given as a function of time t by the expression a equals A sine of omega-t. , whereA. and omega. are constants. What is the period of this motion?

Answer

omega.

2 pi omega.

omega squared times A.

2 pi over omega.

omega over 2 pi.

Well, this question is all about periods and sine functions. But let me clown it up a bit and give you an answer that will make you laugh!

The period of this motion is 2 pi over omega. That's right, it's just like saying "Two pies over omega"! Now that's a tasty mathematical treat, don't you think? So, go ahead and enjoy your delicious period while I go grab a slice of apple pie. Yum!

To find the period of motion for an object moving up and down the y-axis with an acceleration given by a = A*sin(ωt), where A and ω are constants, we can use the formula for the period of a sinusoidal function.

The period, T, is defined as the time it takes for the function to complete one full cycle. In this case, the function is sin(ωt).

The formula for the period of a sinusoidal function is given by:

T = 2π / ω

Therefore, the correct answer is:

2π / ω

To determine the period of the motion, we need to find the time it takes for the object to complete one full cycle of its motion.

The equation given for acceleration is a = A*sin(omega*t), where A and omega are constants. The acceleration is given as a function of time.

The period of a periodic function is the time it takes for the function to repeat itself. In this case, we need to find the value of t at which the sine function repeats.

The period of a sine function is given by the formula T = 2*pi/omega, where omega is the angular frequency of the function.

Comparing this formula with the options given, we can see that the correct answer is:

Period = 2 pi over omega.