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For a simple harmonic motion governed by Hooke's Law, , if is the period then the quantity $\frac{T}{2\pi}$ is equivalent to which of the following?

$\sqrt{\frac{m}{k}}$

$\sqrt{\frac{k}{m}}$

$\frac{m}{k}$

$\frac{k}{m}$

Explanation

We know that T is the period. The equation for T is $T = 2\pi\sqrt{\frac{m}{k}}$ for harmonic motion.

Solve for $\frac{T}{2\pi}$ by dividing the equation by $2\pi$ on both sides. The result is $\sqrt{\frac{m}{k}}$ , which is the answer.

A pendulum is made up of a small mass that hangs on the end of a long string of negligible mass. The pendulum is displaced by and allowed to undergo harmonic motion. What is the angular frequency of the resulting motion?

$g=9.8\frac{m}{s^2}$

$4.4\frac{rad}{s}$

$3.1\frac{rad}{s}$

$19.6\frac{rad}{s}$

$0.044\frac{rad}{s}$

$70\frac{rad}{s}$

Explanation

The angular frequency of a simple pendulum is $\omega = \sqrt{\frac{g}{L}}$ , where is the length of the pendulum.

The bases on a baseball field are 90 feet apart.

A players hits a home run and gets around the bases in 20 seconds what is the players total velocity?

$0\frac{ft}{s}$

$18\frac{ft}{s}$

$20\frac{ft}{s}$

$15\frac{ft}{s}$

$36\frac{ft}{s}$

Explanation

$Velocity = \frac{displacement}{time}$

Displacement is a vector. Magnitude and direction matter.

$Displace = 0 \rightarrow Velocity= 0$

For a simple harmonic motion governed by Hooke's Law, , if is the period then the quantity $\frac{T}{2\pi}$ is equivalent to which of the following?

$\sqrt{\frac{m}{k}}$

$\sqrt{\frac{k}{m}}$

$\frac{m}{k}$

$\frac{k}{m}$

Explanation

We know that T is the period. The equation for T is $T = 2\pi\sqrt{\frac{m}{k}}$ for harmonic motion.

Solve for $\frac{T}{2\pi}$ by dividing the equation by $2\pi$ on both sides. The result is $\sqrt{\frac{m}{k}}$ , which is the answer.

$g=9.8\frac{m}{s^2}$

$4.4\frac{rad}{s}$

$3.1\frac{rad}{s}$

$19.6\frac{rad}{s}$

$0.044\frac{rad}{s}$

$70\frac{rad}{s}$

Explanation

The angular frequency of a simple pendulum is $\omega = \sqrt{\frac{g}{L}}$ , where is the length of the pendulum.

The bases on a baseball field are 90 feet apart.

A players hits a home run and gets around the bases in 20 seconds what is the players total velocity?

$0\frac{ft}{s}$

$18\frac{ft}{s}$

$20\frac{ft}{s}$

$15\frac{ft}{s}$

$36\frac{ft}{s}$

Explanation

$Velocity = \frac{displacement}{time}$

Displacement is a vector. Magnitude and direction matter.

$Displace = 0 \rightarrow Velocity= 0$

The bases on a baseball field are apart.

A player hits a home run and runs around all four bases. What is his total displacement?

Explanation

Displacement is a vector. Therefore, magnitude and direction matters and because direction matters the total displacement is 0 feet.

A projectile reaches it max height in It has a horizontal velocity of $37\frac{m}{s}$ . What is the speed at which it is launched?

$44.65\frac{m}{s}$

$30.65\frac{m}{s}$

$16\frac{m}{s}$

$54\frac{m}{s}$

$37\frac{m}{s}$

Explanation

We first must find the initial vertical velocity ( $v_{iy}$ ) using:

$v_{fy}=v_{iy}+at$

We know that $v_{fy}=0$ , since in 2 dimensions the vertical velocity at its max height is equal to zero.

we also know that because that information is given, and that $a=-g=-10\frac{m}{s^2}$

So plugging in what we know:

$-v_{iy}=-gt$

$v_{iy}=2.5*10=25\frac{m}{s}$

Knowing that $V_{horizontal}$ is constant and equals $37\frac{m}{s}$ , we can use the pythagorean theorem to determine the Resultant initial Velocity:

$V_i^2=V_{ix}^2+v_{iy}^2=37^2+25^2$

$V_i=\sqrt{1994}=44.65\frac{m}{s}$

If the mass of a simple pendulum is quadrupled, then its period __________.

remains the same

quadruples

doubles

is reduce to one quarter

Explanation

We know that the equation for the period of a simple pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$ . This equation does not depend on mass. It is only affected by the length of the pendulum (L) and the gravitational constant (g). Therefore, adding mass to the pendulum will not effect the period, so the period remains the same.

If an object strikes the ground at $37\frac{m}{s}$ what height was it dropped from?

Explanation

First let's make a table of what we know:

$V_{yf}=37\frac{m}{s}$

$a=|g|=10\frac{m}{s^2}$

$v_{yi}=0\frac{m}{s}$ because it has zero velocity right when it is dropped.

It should also be noted it is simpler to define things so that the initial height is zero, so we don't have to deal with a bunch of negative numbers.

Also, since it is dropped vertically and we are only interested in the height it is dropped from we aren't interested in any information regarding the projectile's horizontal motion.

Let's use the equation:

$v{fy}^2=v_{iy}^2+2gh$