# The Physics of Simple Harmonic Motion

An object moving in an SHM must obey the relation

\begin{aligned} a=-\omega ^2 x\end{aligned}

where $x$ is the displacement and $\omega$ is the angular frequency of the oscillator. $\omega$  is related to the frequency $f$ of the oscillator by the relation \begin{aligned} w=2\pi f \end{aligned} .

A body is moving in SHM if its acceleration is directly proportional to the displacement and in the opposite direction to the displacement.

In this equation, we observe the following:

• the acceleration of the oscillator is always opposite to the displacement,
• the magnitude of the acceleration is directly proportional to the displacement.

From these two points, we see that the acceleration is greatest when the oscillator is at the maximum displacement while it is zero when the oscillator is at equilibrium. Since force and acceleration is directly proportional $F=ma$ , we see that the restoring force of the oscillator is correspondingly largest at the maximum displacement and zero at equilibrium.

### SHM in a spring

When a object hung on a spring oscillates, it is moving in SHM. There is a restoring force from the spring causing the object to move up and down. There are two ways to start the motion:

1. displace the object to a certain distance and release, or
2. with the oscillator at equilibrium, give it a push such that it starts moving.

Which colour of the graphs would case 1 and 2 belong?

The red graph would have an equation of \begin{aligned} x=x_0 \sin\omega t \end{aligned} while the blue graph is \begin{aligned} x=x_0 \cos \omega t \end{aligned}.

For the rest of this post, I will use the first equation (red), \begin{aligned} x = x_0 \sin \omega t \end{aligned}.

Since velocity is the derivative of displacement with time,

\begin{aligned} v=\frac{dx}{dt}=x_0 \omega \cos \omega t \end{aligned}

Acceleration is the derivative of velocity with time, hence

\begin{aligned} a=\frac{dv}{dt}= x_0 \omega^2 \sin \omega t\end{aligned}

But from the first equation $x = x_0 \sin \omega t$ , we have

\begin{aligned} a = -x_0 \omega^2 \end{aligned}

### Equation relating velocity to displacement

In the above equations, we have relationships with respect to time. We can also determine the velocity of the object if the displacement is known.

From the original displacement-time equation,

\begin{aligned} x&=x_0 \sin{\omega t} \end{aligned}

\begin{aligned} v=x_0 \omega \cos{\omega t} \end{aligned}

Squaring both equations, we have

\begin{aligned} x^2&=x_0 ^2 \sin^2{\omega t}\\v^2&=x_0 ^2 \omega ^2\cos^2{\omega t}\\&=x_0 ^2 \omega^2 (1-\sin^2{\omega t})\\&=x_0 ^2 \omega^2(1-\frac{x^2}{x_0 ^2})\end{aligned}

Simplifying,

\begin{aligned} v^2 &= \omega^2(x_0 ^2 - x^2) \end{aligned}

Hence,

\begin{aligned} v = \pm\omega \sqrt(x_0 ^2 - x^2) \end{aligned}

### The Restoring Force

It is important to understand that the restoring force is the resultant of both the spring force and the weight of the oscillator. Since weight is constant, the restoring force would be proportional to the spring force, $F=-kx$. Hence, the restoring force is never constant.

\begin{aligned} F_R=F_S-w \end{aligned}

where $F_R$ is the restoring force and $F_S$ is the spring force.

From the equation, it can be seen that the maximum restoring force occurs when $F_S$ is maximum, which occurs when the displacement is at maximum. This is also the position where the acceleration is maximum(and zero velocity).

### Phases of an SHM

\begin{aligned} x &= x_0 \sin \omega t \\ v &= x_0 \omega \cos \omega t\\ a &= -x_0 \omega ^2 \sin \omega t\end{aligned}

As we can see,

• the velocity of SHM $v$ and displacement $x$ has a phase difference of \begin{aligned} \frac{\pi}{2} \end{aligned} .
• The acceleration $a$ and the displacement has a phase difference of $\pi$.
• The acceleration and velocity has a phase difference of $\frac{\pi}{2}$

### Review Question 1

Using the blue graph above, sketch the corresponding graphs of $v$ and $a$.

# Simple Harmonic Motion and Circular Motion

Simple harmonic motion is closely related to a uniform circular motion.

A real life simple harmonic motion.

### Review Question 1

Answer the following questions as you watch the first YouTube video(Simple harmonic motion and uniform circular motion).

1. How does the centre of the red cylinder and the green ball relate?
2. At which part of the SHM(left) is the red cylinder moving the fastest and slowest?
3. At which part of the circular motion(right) is the green ball moving vertically the fastest and slowest?

### Review Question 2

Navigate to this site. This animation shows how a 4-stroke engine internal combustion works.

1. The piston is moving with SHM. At which point is the piston the fastest?
2. At which point is the piston moving the slowest?

Given that the maximum velocity of the piston is $v$,

1. write down the expression of the velocity of the piston at the position shown in the diagram.
2. write down the expression of the acceleration of the piston at the position shown in the diagram.
3. Sketch a velocity-time graph of the SHM demonstrated by the piston. Draw two complete cycles and label the parts A, B and C on your graph.

# Mathematics of Circular Motion

### Angular Displacement

The angle which an object moves around a circle is called the angular displacement. This angle is usually expressed in radians.

\begin{aligned} \pi = 180^\circ \end{aligned}

Hence, there are $2 \pi$ radians in one complete circle.

The relationship between the angle in radian and the circle is

\begin{aligned} \theta =\frac{s}{r} \end{aligned}

where $s$ is the angular displacement in radians, and $r$ is the radius of the circle. The unit of $\omega$ is $\text{rad s}^{-1}$ .

### Angular Velocity

Angular velocity is the rate of change of angular displacement.

Mathematically, angular velocity $\omega$ is

\begin{aligned} \omega = \frac{\Delta \theta}{\Delta t} \end{aligned}

Since an object moves one circumference in one period time, we have

\begin{aligned} \omega &= \frac{2 \pi}{T}\\v &= \frac{2 \pi r}{T}\end{aligned}

Eliminating $T$ ,

\begin{aligned} v=r\omega \end{aligned}

### Review

Question 1

An object moves round a circle of 15 cm radius at a constant speed. It completes one revolution in 8.0 s. Calculate its angular velocity and its velocity.

Solution

\begin{aligned} \omega &= \frac{2 \pi}{8.0}\\&= 0.785\text{ rad s}^{-1}\end{aligned} \begin{aligned} v &= r \omega\\&= 15 \times 0.785\\&= 11.8\text{ cm s}^{-1}\end{aligned}

# How a Capacitor Discharge

### How does a capacitor work?

When a capacitor is connected across a power source, it starts charging up as current flows in the circuit.

Initially when the capacitor has not stored any charge, its potential difference is 0 V. As current flows, charges start storing across its plates, and the potential difference increases. The current also starts to decrease because the capacitor reduces the overall e.m.f of the circuit. When the potential difference of the capacitor equals to the e.m.f. of the power source, current stops flowing.

After the capacitor is fully charged, the switches are set to the figure above. The capacitor starts discharging and a current flows in the left part of the circuit. The potential difference across the capacitor follows an exponential curve.

### Wolfram Demonstration

Click this Wolfram Demonstration Project link to access the simulation. This simulation shows how the potential difference across the capacitor decreases during a discharge. The rate of discharge is dependent on the capacitance of the capacitor. A larger capacitor stores more charge. Although the maximum voltage across it is dependent on the power source, a large capacitor produces a larger current.

Procedure:

1. Adjust all the controls to the left such that d = 0.4, L = 1.5 and time = 0.
1. Observe that the capacitor discharges almost all its stored charge after about 50 s.
2. The capacitor has a capacitance of $0.481\mu \text{F}$ and its maximum stored charge is $21.6 \mu \text{C}$
2. Increase the separation d to 0.7.
1. Observe that the capacitance is now $0.275\mu \text{F}$ and maximum charge $12.4 \mu \text{C}$. This means that increasing the distance between the plates decreases the capacitance, and correspondingly the charge stored, since $Q=CV$ .
2. Lesser stored charge means that the current stops flowing after about 30 s.
3. Increase the length of side to 3.0.
1. Observe that the capacitance is now $1.1 \mu \text{F}$ and that the charge stored is now $49.4 \mu \text{C}$ . Increasing the cross-sectional area increases the capacitance and the stored charge.
2. Since the stored charge is larger, the capacitor takes a longer time to discharge to 0 V.

### Summary

1. Capacitance depends on the physical property. The capacitance is larger is the distance between the plates is small and the cross-sectional is large.
2. A large capacitor stores a larger charge, and takes a longer time to discharge. Conversely, it should take a larger time to charge a large capacitor to its maximum voltage.

# Capacitance

### What is a capacitor?

A capacitor is an electrical component that stores charge. It is usually made by having two parallel plates thinly separated by an insulating material. This material is also known as a dielectric. Current flows through an insulator if the voltage across it is sufficient. In a capacitor, the conducting plates are close enough for the current to flow, even through it is separated by the insulating dielectric. However, because of the dielectric, some charges are stored on the plates as the current flows, resulting in higher potential difference as time passes. When the potential difference across the capacitor equals to the e.m.f of the circuit, current stops flowing.

### Physics of capacitance

Capacitance is defined as

Capacitance is the ratio of the charge stored to potential difference across a capacitor.

Mathematically,

\begin{aligned}C=\frac{Q}{V} \end{aligned}

The SI  unit of capacitance is farad (F).

A capacitor of 1 farad has a stored charge of 1 C when the potential difference across it is 1 V.

### Combined Capacitance

Capacitors can be connected either in series or parallel.

In a series connection, the combined capacitance can be found from the relationship

\begin{aligned} V_\text{total} &= V_1 + V_2 + ...\\\frac{Q}{C_\text{total}} &= \frac{Q}{C_1}+\frac{Q}{C_2}+...\end{aligned}

Since the charge $Q$ stored in each capacitance is the same,

\begin{aligned}\frac{1}{C_\text{total}} &= \frac{1}{C_1}+\frac{1}{C_2}+...\end{aligned}

In a parallel connection, the combined resistance is derived as follow:

\begin{aligned} Q_\text{total}&=Q_1+Q_2+...\\C_\text{total}V&=C_1V+C_2V+...\\\end{aligned}

Since the potential difference $V$ is the same in a parallel circuit,

\begin{aligned} C_\text{total}=C_1+C_2+... \end{aligned}

## Summary

1. Capacitance is the ratio of the charge stored to the potential difference across the component.
2. SI unit of capacitance is farad (F).
3. The total capacitance of series connected capacitors is \begin{aligned} \frac{1}{C_\text{total}}=\frac{1}{C_1}+\frac{1}{C_2}+... \end{aligned}
4. The total capacitance of parallel connected capacitors is \begin{aligned} C_\text{total} = C_1+C_2+... \end{aligned}

## Review

### Question 1

The figure shows three capacitors connected in series with a cell of e.m.f. 3.0 V.

Calculate the p.d. across each capacitor.

### Solution

Since the charge stored in each capacitor is the same as the charge stored across all the capacitors, we shall go ahead and find the charge stored in the combined capacitors and then use it to calculate the potential difference across each capacitor.

\begin{aligned} \frac{1}{C_\text{total}} &= \frac{1}{100 \times 10^{-3}}+\frac{1}{200 \times10^{-3}} + \frac{1}{400 \times 10^{-3}}\\C_\text{total}&= 0.0571\text{ F}\\ Q_\text{total}&=3.0 \times 0.0571\\&=0.1713\text{ C}\end{aligned}

Hence,

\begin{aligned} V_1&=\frac{0.1713}{100 \times 10^{-3}}\\&=1.713\text{ V}\\V_2&=\frac{0.1713}{200 \times 10^{-3}}\\&=0.857\text{ V} \\V_3&=\frac{0.1713}{400 \times 10^{-3}}\\&=0.428\text{ V}\end{aligned}

You can observe that the sum of all the three p.d. is the e.m.f. of the cell.

# Gravitation and Circular Motion

### Why does a orbiting satellite not fall to the Earth?

A common misconception among students is that orbiting satellites do not experience gravity. It is argued that since these satellites are not falling to the Earth, they must experience no gravity. This thought is not true because the very action of orbiting around the Earth shows that the satellite is under the influence of Earth’s gravity. Then why is the satellite not falling to the Earth?

The below simulation explains why satellites are not falling back to the Earth. In fact, these satellites are actually in the process of falling to the Earth. The simulation explains clearly how this works.

Satellite orbiting the Earth

So the answer to the above question is that the satellite is in fact, falling to the Earth. However, the Earth’s surface curves away at the same rate the the satellite is falling towards Earth. Hence, the satellite never falls onto the Earth.

It is important to note that this can happen because the satellite possesses an initial velocity that is parallel to the Earth’s surface. Without sufficient velocity of this component, the satellite would fall towards Earth.

### Gravitational force and circular motion

Planets revolve around the Sun and the Moon revoles around the Earth. These orbits are generally approximately circular, and the centripetal force to pull them around the orbits is the gravitational force.

\begin{aligned} \frac{GMm}{r^2} &= \frac{mv^2}{r} \\ \frac{GM}{r} &= {v^2}\\v&=\sqrt{\frac{GM}{r}}\end{aligned}

We can see that the linear speed, hence the angular speed, of an orbiting body is independent of the mass of the body. It is only dependent on the mass of the body producing the gravitational field and the distance of the orbiting body from the centre of the main mass.

To find the relationship between the radius and the period of orbit,

\begin{aligned} (\frac{2\pi r}{T})^2 &= \frac{GM}{r}\\T^2 &= \frac{4\pi^2}{GM}r^3 \end{aligned}

Hence we can see that the square of the period is proportional to the cube of the radius of orbit. The is the Kepler’s third law.

### Summary

1. When an object orbits around another mass, the centripetal force is gravitational in nature.

2. The square of the period is proportional to the cube of the radius of orbit. This is also known as the Kepler’s third law.

### Review

#### Question 1

Two stars orbit each other in a time of 3.0 years. Calculate the angular speed \begin{aligned} \omega \end{aligned} for each star. Given that the ratio \begin{aligned} \frac{M_1}{M_2}=2.0 \end{aligned} , and the separation of the starts is $3.0 \times 10^{12} \text{ m}$ , calculate the radii \begin{aligned} R_1 \end{aligned} and \begin{aligned} R_2 \end{aligned} .

#### Solution:

Angular speed is the angular displacement per unit time. Hence,

\begin{aligned}\omega &= \frac{2 \pi}{3.0 \times 365 \times 24 \times 3600}\\&=6.6\times 10^{-8}\text{rad s}^{-1} \end{aligned}

The two stars must be orbiting with the same angular speed. Otherwise, there will be a point in time when the inner star catches up and the centre of mass would change position. The gravitational force between the two stars would then change, changing the centripetal force. This is not possible for a stable star system, hence the angular speeds of the two stars must be the same.

Equating the gravitational force to the centripetal force,

\begin{aligned} \frac{GM_1M_2}{(R_1 + R_2)^2 } &= M_2 R_2 \omega^2\\\frac{GM_1}{(R_1 + R_2)^2} &= R_2 \omega^2 \end{aligned}

Similarly,

\begin{aligned} \frac{GM_2}{(R_1 + R_2)^2} &= R_1 \omega^2 \\\frac{M_1}{M_2} &= \frac{R_2}{R_1}\end{aligned}

Since \begin{aligned} \frac{M_1}{M_2} = 2.0 \end{aligned} ,

\begin{aligned} \frac{R_2}{R_1}&=2.0\\R_2&=2.0R_1\\R_1+2R_1 &= 3.0 \times 10^{12}\\R_1&=1.0\times 10^{12}\text{ m}\\R_2 &= 2.0 \times 10^{12} \text{ m} \end{aligned}

# Gravitation – Potential and Potential Energy

## Gravitational Potential and Potential Energy

Gravitational potential at a point is the amount of work needed to bring a unit mass from infinity to that point.

Gravitational potential is defined to be negative since gravity is attractive in nature. A negative potential means that no external work is needed to bring a unit mass from infinity to that point, since the gravity-producing mass would be doing the work to pull the unit mass to that point.

Mathematically,

\begin{aligned} \phi = -\frac{GM}{r} \end{aligned}

Note that potential is a scalar quantity. Hence, the potential at a point is simply the algebraic sum of the potential of the different masses at that point.

\begin{aligned} \phi_\text{sum} = -\frac{GM_1}{r_1} - \frac{GM_2}{r_2} - ... \end{aligned}

Similarly, gravitational potential energy is defined as

Gravitational potential energy of a mass at a point is the amount of work needed to bring the mass from infinity to that point.

\begin{aligned} \text{GPE} = -\frac{GMm}{r^2} \end{aligned}

You may find this concept similar to gravitational force and field strength. Both gravitational force and potential energy invloves the product of two masses \begin{aligned} Mm \end{aligned} while field strength and potential involves just the gravity-producing mass \begin{aligned} M \end{aligned}.

## Gravitational Potential vs Field Strength

It is easy to compare the relative values of potential and field strength because of the similar form of equations.

\begin{aligned} g &= -\frac{GM}{r^2} \\ \phi &= -\frac{GM}{r}\end{aligned}

## GPE = mgh

In many junior Physics text, gravitational potential energy is quoted with the formula
\begin{aligned} \text{GPE} = mgh \end{aligned}
This formula assumes that the change in height in insignificant compare to the radius of the Earth. This formula also calculates the change in the potential energy due to a change in position.

The formula \begin{aligned} F_G = \frac{GMm}{r} \end{aligned} calculates the actual amount of potential energy a mass possess due to its position. This formula does not calculate the change in potential energy. To calculate the change in potential energy,

\begin{aligned}\text{change in GPE} &= \frac{GMm}{r_1} - \frac{GMm}{r_2}\\\end{aligned}

If \begin{aligned} r_1 \approx r_2 \end{aligned} ,

\begin{aligned} \text{change in GPE} &= GMm ( \frac{1}{r_1} - \frac{1}{r_2} ) \\&= GMm (\frac{r_2-r_1}{r_1 r_2})\\&=gm(r_2-r_1)\\&=mgh\end{aligned}

From this, we have our old formula \begin{aligned} \text{GPE} = mgh \end{aligned}

### Summary

1. Gravitational potential at a point is the amount of work needed to bring a unit mass from infinity to that point.
2. Gravitational potential energy of a mass at a point is the amount of work needed to bring the mass from infinity to that point.
3. \begin{aligned} \phi = -\frac{GM}{r} \end{aligned}
4. The gravitational potential has a larger magnitude than the field strength.
5. When the change in height is small, we may use \begin{aligned} \text{change in GPE} = mgh \end{aligned}

# Gravitation – Force and Field Strength

## Concept of a Gravitational Field and Force

A gravitational field is a region in which a mass experiences a force.

A mass, \begin{aligned} m \end{aligned} that is present inside a gravitational field experiences a gravitational force. This gravitational field is produced by another mass \begin{aligned} M \end{aligned} .

The amount of force experienced by the mass \begin{aligned} m \end{aligned} is directly proportional to the product of the two masses and indirectly proportional the the square of the separation.

\begin{aligned} F_G=\frac{GM m}{r^2} \end{aligned}

The separation \begin{aligned} r \end{aligned} is the distance between the centre of mass of the two masses. We often assume that the two masses are point masses if the separation is large relative to the radius of the masses. If the separation is not large, then it is important to use the distance between the centre of mass of the two masses. We should not use the separation between the two surfaces of the masses. Furthermore, one may safely assume that for a uniform sphere, the centre of mass is the centre of the sphere.

There are two regions about the field produced by a mass: the region outside the mass and the region inside. The region outside the mass follows the inverse exponential relationship of 1/r^2.

Inside the mass, the relation is direct proportional to the distance from the centre of the mass. This is because as you proceed nearer to the centre of the mass, there is less mass “below” you. The part of the mass “above” you pulling you “up” is offset by the mass “below” you pulling you down.

### Gravitational Field Strength

Gravitational field strength is often misunderstood. Its definition is

Gravitational field strength at a point is the gravitational force acting on a unit mass at that point.

From the definition,

\begin{aligned} \text{force} &= m \times g\\mg &= \frac{GMm}{r^2}\\g &= \frac{GM}{r^2}\end{aligned}

We can observe that gravitational field strength only depends on the gravity-producing mass and the distance from it. It is not dependent on the test mass.

There is only one value of gravitational field strength at any particular point since we are always comparing the gravitational force on one unit mass. If there are multiple masses creating gravitational fields, the gravitational field strength at any particular point would be the vector sum of all the field strengths due to the the different masses.

### Field Strength on the surface of the Earth

When we calculate weight of an object, we always use the formula \begin{aligned} w = mg \end{aligned}

\begin{aligned} g \end{aligned} is referred to as the gravitational field strength (although it is commonly stated as the gravitational acceleration). Near the Earth’s surface, the field strength of the Earth is

\begin{aligned} g &=G\frac{M}{r_\text{Earth radius}^2}\\&=G\frac{M}{6400000^2}\end{aligned}

The field strength at 10 km above the surface would be \begin{aligned} G\frac{M}{6410000^2} \end{aligned}.

The difference among them is negligible. Hence we assume that the gravitational field strength on Earth’s surface is constant.

### Activity 1

Access the online PHET simulation.

Objective:
What is the relationship between the force acting on m2 by m1 and the force of m1 on m2?

1. Using the default values, observe the force on m2 by m1 and the force on m1 by m2.
What do you observed?
2. Change the mass of m1 to another value.
What do you observed?
3. Explain.

### Activity 2

Access the same simulation as activity 1. Also open this activity sheet to download a copy of the activity on iCloud.

Objective:
What is the relationship between the field strength with distance?

1. Change m2 to 1 kg, while m1 remains as 50 kg.
2. Drag m2 to the 10 m mark, and m1 to the 0 m mark. You may not be able to do this exactly, but an approximate position would be fine.
3. Record the force on m2 by m1 in the table.
4. Repeat this for distances 9, 8, 7, 6, 5, 4, 3, 2 and 1 m.
5. Observe the shape of the graph.

Question:

1. Will the gravitational field strength ever reaches zero?
2. At which point(from 0 m to infinity) is the field strength the strongest?

### Summary

1. understand the concept of a gravitational field as an example of a field of force and define gravitational field strength as force per unit mass.
2. understand that, for a point outside a uniform sphere, the mass of the sphere may be considered to be a point mass at its centre.
3. recall and use Newton’s law of gravitation in the form \begin{aligned} F=G\frac{Mm}{r^2} \end{aligned}

### Review

##### Question 1

Geostationary satellites are satellites that orbit around the Earth with a period of 24 hours. The satellite would appear as a stationary point relative to an observer on Earth. Calculate the distance above the Earth’s surface of a geostationary satellite. Properties of Earth

##### Solution

Since the centripetal force of the satellite is due to gravity,

\begin{aligned} \frac{GMm}{r^2} &= mr\omega ^2\\\frac{GM}{r^2} &= r \omega ^2 \\r &= \sqrt[3]{\frac{GM}{w}} \end{aligned}

$\omega$ is known, since the geostationary satellite must make one orbit in one day,

\begin{aligned} \omega &= \frac{2 \pi}{24 \times 3600}\\&= 7.3\times 10^{-5} \text{ rad s}^{-1}\\r&=42.1\times 10^6\text{ m}\end{aligned}

Hence, the distance above Earth’s surface is

\begin{aligned} d &= (42.1 - 6.38) \times 10^6 \text{ m}\\ &= 35.7 \times 10^3 \text{ km} \end{aligned}

It is interesting to note that the distance of Moon from Earth is about $370 \times 10^3 \text{ km}$. Hence, a geostationary satellite is about 10% of the distance from Earth to the Moon.

# Work done in a Capacitor

Capacitor is an electrical component that stores charge. As charges are stored, potential energy in the capacitor also increases. In this post, I would like to explain how to calculate the energy stored in a capacitor.

##### Question

The switch is closed at time t = 0 s. During the time interval from t = 0.0 s to t = 4.0 s, 15 mC charge passes through the resistor.

Calculate the energy transferred by the battery during this 4.0 seconds and the energy stored in the capacitor after 4.0 s.

#### Definition of Potential Difference

In many textbooks, we learnt that the potential difference across a component is the energy released for every unit charge passing through it.

$V=\frac{W}{Q}$

If we are to calculate the energy released by the battery, it would be

$W=V \times Q = 9.0 \times 15 \times 10^{-3} =0.135 \text{J}$

#### Energy stored in the capacitor

However, the energy stored in the capacitor cannot be 0.135 J. This is because a capacitor’s potential difference changes as it stores charge. The equation above requires that the p.d. remains constant. Hence, we need to use another formula.

Since capacitance is defined as the charge stored for every unit potential difference,

$Q=VC$

i.e. the charge stored is directly proportional to the potential difference across it. Since the work done is the area under a charge-voltage graph, we will use the formula for a triangle.

$W=\frac{1}{2}QV$

At time t = 4.0 s,

$V=\frac{15 \times 10^{-3}}{2000 \times 10^{-6}}=7.5 \text{V}$

Hence,

$W=\frac{1}{2} \times 15 \times 10^{-3} \times 7.5 = 0.05625 \text{J}$

#### Lost energy?

The energy stored in the capacitor(0.05625 J) is not the same as the energy transferred from the battery(0.135 J) because heat is lost through the resistor as current flows.

# Ultrasound Calculations

The table provides the data for the acoustic impedance and absorption coefficients for muscle and bone.

A parallel beam of ultrasound of intensity $I_0$ enters the muscle of thickness 3.0 cm as shown. The ultrasound is then reflected at the tissue boundary and returns to the surface of the muscle.

Calculate the intensity, in terms of $I_0$, that is received when the ultrasound returns back to the surface of the muscle.

There are three important parts to solve this problem. The first part would be the attenuation of intensity inside the muscle. The second part would be the reflection of ultrasound at the muscle-bone boundary. The last part would be the intensity attenuation inside the muscle back to the surface.

$I=I_0 e^{-21 times 0.03}$