All Solutions
Page 491: Standardized Test Prep
We know that period $T$ is defined as the time required to complete one full cycle of a given motion.
The frequency $f$ is the number of oscillations per unit time.
They are related by the following formula:
$$
begin{equation}
f = frac{1}{T}
end{equation}
$$
The time that passes between two adjacent nodes is one half-period. From the figure we see that this time is $1.5text{ s}$. So we conclude the period is:
$$
begin{align*}
T = 2 cdot 1.5text{ s} = 3text{ s}
end{align*}
$$
Now to find the frequency we use formula (1) directly:
$$
begin{align*}
f &= frac{1}{T} \
f &= frac{1}{3text{ s}} approx 0.33text{ Hz}
end{align*}
$$
The distance from the maximum elongation to the equilibrium position is the amplitude distance. From the figure we read this to be
$$
begin{align*}
A = 2text{ m}
end{align*}
$$
So se read from the figure the times when the elongation is zero.
The times are:
$$
begin{align*}
t = 1.5text{ s}text{ , } t = 3text{ s} text{ , } t = 4.5text{ s}
end{align*}
$$
So the correct answer is B
A mass on a spring is an oscillator, so the conclusion holds. The speed is maximum in the equilibrium position, when both the displacement and force are minimum, that is zero.
So the correct answer is C.
This ratio is nicely described by the sine function, where the argument is the angle of displacement from the equilibrium line. For small angles the sine is the same as the argument itself:
$$
begin{gather*}
text{For } theta ll 1 (text{in radians}): \
sintheta approx theta
end{gather*}
$$
So we come to the conclusion that the correct answer is C.
We know that when a standing wave is formed in a string the number of antinodes determines the harmonic of the wave. This gives us a result that the length of the string is equal to a integer number of wavelength halves, that is:
$$
begin{align*}
L = frac{n cdot lambda}{2}
end{align*}
$$
Rearranging:
$$
begin{gather*}
2 cdot L = n cdot lambda \
lambda = frac{2L}{n}tag{1}
end{gather*}
$$
From this we see that any wavelength lambda that fulfils this equation for an integer n can produce a standing wave.
The first harmonic is the one with maximal wavelength. Knowing this we set the integer $n = 1$, and from formula (1) get:
$$
begin{align*}
lambda = 2 cdot L
end{align*}
$$
This result symbolically represents statement B, which we now conclude is the correct statement.
All the other statements are either false, or correct only in some situations, not generally.
We know that period $T$ is defined as the time required to complete one full cycle of a given motion.
The frequency $f$ is the number of oscillations per unit time.
They are related by the following formula:
$$
begin{equation}
f = frac{1}{T}
end{equation}
$$
We know that when a standing wave is formed in a string the number of antinodes determines the harmonic of the wave. This gives us a result that the length of the string is equal to a integer number of wavelength halves, that is:
$$
begin{align*}
L = frac{n cdot lambda}{2} \
end{align*}
$$
Rearranging:
$$
begin{gather*}
2 cdot L = n cdot lambda \
lambda = frac{2L}{n}tag{2}
end{gather*}
$$
From this we see that any wavelength lambda that fulfils this equation for an integer $n$ can produce a standing wave.
We know that the wavelength and frequency determine the speed of a wave. They are related by the following formula:
$$
begin{equation*}
v = lambda cdot f tag{3}
end{equation*}
$$
Since we know the frequency to be equal $f = 2.4text{ Hz}$, the period of the wave can easily be found using the well known formula (1) as follows:
$$
begin{align*}
& f = frac{1}{T} \
& T = frac{1}{f} = frac{1}{2.4text{ Hz}} approx 0.42text{ s}
end{align*}
$$
Now for the wavelength, we know the first harmonic is the one with maximal wavelength. Knowing this we set the integer $n = 1$, and from formula (2) get:
$$
begin{align*}
lambda = 2 cdot L
end{align*}
$$
Using the value for the length $L = 0.75text{ m}$ we get:
$$
begin{align*}
lambda = 2 cdot 0.75text{ m} = 1.5text{ m}
end{align*}
$$
Finally, to find the speed of the waves on the string we use formula (3) and the value we just calculated for the wavelength $lambda$ , along with the given value for the frequency $f$:
$$
begin{align*}
v = f cdot lambda
end{align*}
$$
Plugging in the numbers we get:
$$
begin{align*}
v = 2.4text{ Hz} cdot 1.5text{ m} = 3.6 ; frac{text{m}}{text{s}}
end{align*}
$$
The correct solution is A, the period is $T = 0.42text{ s}$ , the speed is $v = 3.6 ; frac{text{m}}{text{s}}$
We know that the wavelength and frequency determine the speed of a wave. They are related by the following formula:
$$
begin{equation}
v = lambda cdot f
end{equation}
$$
When a wave travels from one medium to another the frequency is conserved, so we write:
$$
begin{align*}
f_1 = f_2
end{align*}
$$
We know from the problem that the wave travels from a medium where its speed is greater, to a medium where its speed is lower, so we can write:
$$
begin{align*}
v_2 < v_1
end{align*}
$$
Now to find how the wavelength changes we use formula (1):
$$
begin{align*}
lambda_1 &= frac{v_1}{f_1} \
lambda_2 &= frac{v_2}{f_2}
end{align*}
$$
Since the frequencies are the same, and the wavelength is proportional to the speed the medium with the higher speed will have the higher wavelength.
So we conclude that:
$$
begin{align*}
lambda_2 < lambda_1
end{align*}
$$
So the wavelength decreases.
The goal is to measure the gravitational constant g using a simple experiment with an oscillator and standard measuring equipment.
The first idea that comes to mind, is using a simple pendulum along with a device for measuring length, and a stopwatch for measuring the period of oscillation. As we well know, the period of a pendulum oscillating due to gravitational acceleration $g$ can be found using the length of the pendulum $L$ as follows:
$$
begin{equation}
T = 2 , pi sqrt{frac{L}{g}}
end{equation}
$$
So, by manipulating the length of the pendulum $L$, and then measuring the period $T$, one can find the gravitational acceleration $g$ as follows:
$$
begin{align*}
T^2 &= 4 , pi^2 ; frac{L}{g} \
g &= frac{4 , pi^2 , L}{T^2}
end{align*}
$$
The second method of accomplishing our given task is using a spring mass system, let us first remember the underlying theory behind the method.
For a mass $m$ attached to a spring with a spring constant $k$ we know that the period of simple harmonic oscillations will be:
$$
begin{equation*}
T = 2 , pi ; sqrt{frac{m}{k}} tag{2}
end{equation*}
$$
When a spring is compressed or pulled by a given force $F$ its elongation $x$ is proportional to that force, and the coefficient is called the spring constant.
This is described by Hooke’s Law:
$$
begin{equation*}
F = k , x tag{3}
end{equation*}
$$
$$
begin{align*}
T = 2 , pi ; sqrt{frac{m}{k}}
end{align*}
$$
Rearranging for the ratio $frac{m}{k}$ we get:
$$
begin{align*}
T^2 &= 4 , pi^2 ; frac{m}{k} \
frac{m}{k} &= frac{T^2}{4 , pi^2}
end{align*}
$$
This ratio will now be used to find $g$.
Next we let the spring extend and reach equilibrium in vertical position. This stretching is governed by Hooke’s law, so we will use formula (3). In our case the force $F$ stretching the spring is the weight of the attached mass $m$.
So we can write:
$$
begin{align*}
m , g = k , x
end{align*}
$$
Rearranging for $g$ we get:
$$
begin{align*}
g = frac{k}{m} cdot x
end{align*}
$$
Since we have found the ratio $frac{m}{k}$, we just need to measure the elongation $x$ which can be done using using a ruler or measuring tape.
If we need more than one measurement we can modify the parameters $m$ and $k$, by changing the object or the spring. This change will effect the measured $T$, and through it will be reflected in the final value for $g$.
By using a spring mass system, and changing the spring or the mass.
