For this question, we need to recall what simple harmonic motion is. Remember that it is a periodic motion where the restoring force depends on the displacement of the object undergoing these motions. So to answer this question, we need to keep this idea in mind and see which example doesn't match up.

A mass on a pendulum moving back and forth is clearly an example of simple harmonic motion. As the mass moves further from the center in either direction, it experiences a greater and greater force in the opposite direction.

A child swinging on a swing set is another correct example. This situation is analogous to the mass on a pendulum swinging back and forth.

A vibrating guitar string is yet another example of simple harmonic motion. After it is plucked, the string oscillates back and forth.

Finally, a book falling to the ground does not represent harmonic motion. Once the book is released from rest, we intuitively know that it will fall to the ground and will then stay there; in no way is there any periodic motion.

We know that the equation for the period of a simple pendulum is \(\displaystyle T = 2\pi\sqrt{\frac{L}{g}}\). 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.

Since we are solving for string tension, we need to use the frequency equation with the tension variable in it. That equation is \(\displaystyle f = \frac{n}{2L}\sqrt\frac{T}{\mu}\) where \(\displaystyle f\) is the frequency, \(\displaystyle n\) is the number of the harmonic, \(\displaystyle L\) is the length of the string, \(\displaystyle \mu\) is the linear density of the string, and \(\displaystyle T\) is the tension of the string.

We are given:

\(\displaystyle n=2\)

\(\displaystyle L=37.0 cm\)

\(\displaystyle f=770 Hz\)

\(\displaystyle \mu=2.9\frac{g}{m}\)

Next we must convert the length of \(\displaystyle 37.0 cm\) to meters which is \(\displaystyle 0.37m\) and the mass density of \(\displaystyle 2.9\frac{g}{m}\) to \(\displaystyle 0.0029\frac{kg}{m}\). Then we plug in our known values into the equation and solve for the string tension. The result is \(\displaystyle 373N\).

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

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

First, we need to find the length of the pendulum. Begin with the equation for finding the period of a pendulum:

\(\displaystyle T = 2\pi \sqrt\frac{L}{g}\)

solve for \(\displaystyle L\) to get:

\(\displaystyle L = \frac{T^2g}{4\pi^2}\)

Now we can plug in our given values:

\(\displaystyle L = \frac{(1.82)^2(9.81)}{4\pi^2} = 0.823m\)

Since we have the length of the pendulum determined, we can now find the period of the pendulum on Mars:

\(\displaystyle T = 2\pi\sqrt\frac{L}{g} = 2\pi \sqrt\frac{0.823}{3.71}= 2.96 s\)

\(\displaystyle speed= \frac{distance}{time}\)

\(\displaystyle Distance = 360ft\)

\(\displaystyle Time= 20s\)

\(\displaystyle \frac{360ft}{20s}= 18\frac{ft}{s}\)

There is 90 feet between each base and distance does not depend on direction. Distance is scaler therefore, the batter travels 360 feet.

\(\displaystyle Velocity = \frac{displacement}{time}\)

Displacement is a vector. Magnitude and direction matter.

\(\displaystyle Displace = 0 \rightarrow Velocity= 0\)

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

We are dealing with motion in one plane and the initial velocity is zero. We must use the following kinematics equation to plug in known values and solve:

\(\displaystyle d = \frac{1}{2}at^2\)

\(\displaystyle d = \frac{1}{2}(10\frac{m}{s^2})(4s^2)\)

\(\displaystyle d = 80m\)