4.101 Speed of car – Doppler effect

4.101 Speed of car – Doppler effect

Problem 4.101: A policeman on duty detects a drop of $30\: \%$ in the pitch of the horn of a motor-car as it crosses him. Calculate the speed of car, if the velocity of sound is $330m/s$.

Solution:

How to solve this problem? First try to visualize….There is a car and a policeman. The car is honking its horn. The policeman naturally hears the sound. But the car is moving towards him. He will hear an increase in pitch or frequency. This is due to Doppler effect. Now the car will go past him and will move away from him. He hears the sound with a lower frequency again due to Doppler effect.

The problem says  there is a drop of $30\%$ in the pitch. This is the percentage difference between the high frequency heard while the car approached him and the lower frequency heard while the car move past him.

Let $f$ be the actual frequency of the sound. Policeman hear this as frequency $f_1$ as th car moves towards him and as $f_2$ as the car moves away from him. Then second frequency is $30\:\%$ less than the first frequency. It is only $70\:\%$ of $f_1$.

Mathematically, $f_2$ =  $70\:\%$ of $f_1$.

Or, $f_2$ = $\frac{70}{100}\:f_1$.

Then, $\frac{f_2}{f_1}\:=\:\frac{70}{100}\:=\:\frac{7}{10}$------$(1)$.

Now we need to calculate $f_1$ and $f_2$ separately.

Calculation of $f_1$.

In Doppler effect the apparent frequency $f_1$ of a source moving towards a stationary observer is given by , $f_1\:=\:f\frac{V}{V\:-\:V_s}$------$(2)$.

Calculation of $f_2$.

The apparent frequency $f_2$ of a source moving away a stationary observer is given by , $f_2\:=\:f\frac{V}{V\:+\:V_s}$------$(3)$.

From $(1)$, $(2)$ and $(3)$,

$\frac{f_2}{f_1}\:=\:\frac{f\frac{V}{V\:+\:V_s}}{f\frac{V}{V\:-\:V_s}}\:=\:\frac{7}{10}$

Or, $\frac{V\:-\:V_s}{V\:+\:V_s}\:=\:\frac{7}{10}$.

Then, $10V\:-\:10V_s\:=\:7V\:+\:7V_s$.

Or, $7V_s\:+\:10V_s\:=\:10V\:-\:7V$.

That is, $17V_s\:=\:3V$

From this, $V_s\:=\:\frac{3V}{17}$.

Given that speed of sound in air is $330\:ms^{-1}$.

Therefore, speed of car,

$V_s\:=\:\frac{3\times 330}{17}\:=\:\frac{990}{17}\:=\:58.24\:ms^{-1}$.

So the car was traveling with a speed of $58.24\:ms^{-1}$.

4.100 Maximum acceleration in simple harmonic motion

4.100: A mass of $4\: kg$ suspended from a spring of force constant $800 \:Nm^{-1}$ executes simple harmonic oscillations.   If the total energy of the oscillator is $4\:J$, determine the maximum acceleration in $ ms^{-2}$  of the mass.

Solution:

Angular frequency of spring of constant $k$ and mass $m$ is $\omega\:=\:\sqrt{\frac{k}{m}}$

Here for the given system$\omega\:=\:\sqrt{\frac{800}{4}}\:=\:10\sqrt{2}$.

In simple harmonic motion total energy is $\frac{1}{2}m\omega^2A^2$

Therefore $\frac{1}{2}\times 4\times (10\sqrt{2})^2A^2\:=\:4$.

Solving we get amplitude $A\:=\:0.1\:m$.

Maximum acceleration is $a_{max}\:=\:omega^2 A$.

So, $a_{max}\:=\:(10\sqrt{2}) ^2 \times 0.1A\:=\:20\:ms^{-2}$

4.99 Maximum speed in simple harmonic motion

4.99 Maximum speed in simple harmonic motion

 A spring of force constant $1200 \:N m^{-1}$ is mounted on a horizontal table as shown in figure. A mass of $3 \:kg$ is attached to its free end and pulled sideways to a distance of $2\: cm$ and released. Calculate the maximum velocity of the mass.

 15 -01-2015

Solution:

When the mass is displaced through $2\: cm$, the spring also elongates through the same distance.

Work done by the applied force is stored as elastic potential energy in the spring.

Elastic potential energy in a spring of constant $k$ compressed or elongated through a distance $x$ is given by $\frac{1}{2}kx^2$. When the applied force is removed the spring pulls the mass to the initial position and tries to come to the initial unstretched state.

The elongation of the spring will decrease and the speed of the block will increase.

The energy stored in the spring decreases and is transferred to the block in the form of kinetic energy.

Kinetic energy will be maximum when the spring reaches the unstretched state.

If $V_{m}$ is the maximum speed of the block, then the maximum kinetic energy will be $\frac{1}{2}mv_{m}^2$.

Assuming no energy loss, we can equate the initial energy in the spring to the maximum kinetic energy of the block.

So, $\frac{1}{2}mv_{m}^2\:=\:\frac{1}{2}kx^2$.

Or, $mv_{m}^2\:=\:kx^2$.

$v_{m}^2\:=\:\frac{kx^2}{m}$.

Maximum speed of the block, $v_{m}\:=\:\sqrt{\frac{kx^2}{m}}$.

So the maximum speed of the block will be $\sqrt{\frac{k}{m}}x$.

Using mass, $m\:=\:3\:kg$, spring constant, $k\:=\:1200\:Nm^{-1}$ and displacement, $x\:=\:2\:cm\:=\:0.02\:m$.

Maximum speed of the block will be $\sqrt{\frac{1200}{3}}\times 0.02\:=\:8\:ms^{-1}$.

 

4.98 Length of a pendulum

4.98 Length of a pendulum

Problem 4.98: A simple pendulum is released at rest while its string is making an angle of $10^o$ with the vertical. $0.2$ seconds later the string makes an angle of $5^o$ with the vertical. Determine the length of the pendulum. Take acceleration due to gravity as $10\:ms^{-1}$.

14-Jan-2015 Simple pendulum

Solution:

As the angles are small the pendulum is in simple harmonic motion.

When released as given in the problem, the pendulum will execute simple harmonic

The angular displacement of the pendulum can be written as $\theta\:=\:\theta_{o}\:cos\:\omega t$.

Here $\theta$ is the angle made at time $t$ and $\theta_o$ is the maximum angle or amplitude. These angles must be in radian.

$\omega$ is the angular frequency which is equal to $\sqrt{\frac{g}{L}}$.

Given that initial angle is $10^o$ which is $10\times \frac{\pi}{180}\:=\:\frac{\pi}{18}$ radian.

At a later time $t\:=\:0.2\:s$, angle is $5^o$ which is $5\times \frac{\pi}{180}\:=\:\frac{\pi}{36}$ radian.

Therefore, $\frac{\pi}{36}\:=\:\frac{\pi}{18}\:cos\:\sqrt{\frac{g}{L}} t$.

So, $\frac{1}{2}\:=\:cos\:\sqrt{\frac{g}{L}} \times 0.2$.

As $cos \:\frac{\pi}{3}\:=\:\frac{1}{2}$
we get, $\sqrt{\frac{g}{L}} \times 0.2\:=\:\frac{\pi}{3}$.

Or, $\sqrt{\frac{g}{L}} \times 0.2\:=\:\frac{\pi}{3}$.

Or, $\sqrt{\frac{g}{L}} \:=\:\frac{\pi}{3\times 0.2}$.

Or, $\sqrt{\frac{g}{L}} \:=\:\frac{3.14}{0.6}\:=\:5.33$.

Squaring the equation, we get, $\frac{g}{L} \:=\:27.38$.

Using $g\:=\:10\:ms^{-2}$, $\frac{10}{L} \:=\:27.38$.

From this ,lemgth of the pendulum, $L\:=\:\frac{10}{27.38}\:=\:0.36\:m$.

So the length of the pendulum will be $0.36\:m$.

4.97 Time period of oscillation

4.97 Time period of oscillation

Problem 4.97: A block of mass $M$ kept on a smooth horizontal surface is connected to three identical springs each with constant $k$. Determine the time period for small amplitude oscillations. Neglect all dissipating forces.13-Jan-2015 Oscillations  A

Solution:

When the block is displaced to one side and then released, it will start oscillating. The oscillations will be simple harmonic if acceleration is proportional and opposite to the displacement of the block.

When the block is displaced to the right side through a distance $x$ the two springs on the left side will elongate through $x$ and the springs on the right side will get compressed through the same distance.13-Jan-2015 Oscillations  B 

The forces developed in the springs will be as shown in the diagram.13-Jan-2015 Oscillations  C

The net force on the block is $3kx$.

These forces are opposite in direction of the displacement of the block.

So in vector form it can be written as $\vec{F}\:=\:-\:3k\vec{x}$.

Acceleration of the block, $\vec{a}\:=\:\frac{\vec{F}}{M}\:=\:\frac{-\:3k\vec{x}}{M}$.

It can be seen that the acceleration of the block is proportional and opposite of the displacement. Therefore the motion of the block is simple harmonic in nature.

In simple harmonic motion, acceleration is given by $-\:\omega^\vec{x}$ where $\omega$ is the angular frequency of oscillation.

Therefore, $-\:\omega^2 \vec{x}\:=\:\frac{-\:3k\vec{x}}{M}$.

Therefore, $\omega^2\:=\:\frac{3k}{M}$.

Or, angular frequency, $\omega\:=\:\sqrt{\frac{3k}{M}}$.

Time period $T$ is related to angular frequency as $T\:=\:\frac{2\pi}{\omega}\:=\:\frac{2\pi}{\sqrt{\frac{3k}{M}}}$

From this time period, $T\:=\:2\pi\sqrt{\frac{M}{3k}}$.

4.96 Amplitude of oscillation of a spring

4.96 Amplitude of oscillation of a spring

Problem 4.96: A block of mass $M$ is connected to a series connection of two springs with spring constants $k_1$ and $k_2$ respectively. This is shown in the diagram. If the block oscillates in a simple harmonic fashion with amplitude determine the amplitude of oscillation of the point P.

12-Jan-2015 Oscillations  A

Solution:

Let us assume that the block is displaced through a distance $x$ using a force $F$ as shown. Both the springs will elongate under the force.

Let $x_1$ and $x_2$ be the elongations in the two springs.

Force developed in the springs will be $k_1x_1$ and $k_2x_2$.

The block and the springs are in equilibrium. Therefore $F\:=\:k_1x_1\:=\:k_2x_2$.

$k_1x_1\:=\:k_2x_2$.12-Jan-2015 Oscillations  B

Or, $\frac{k_1}{k_2}\:=\:\frac{x_2}{x_1}$.

Or, $\frac{k_1}{k_2}+1\:=\:\frac{x_2}{x_1}+1$.

Or, $\frac{k_1+k_2}{k_2}\:=\:\frac{x_2+x_1}{x_1}$.

The Displacement of the block must be equal to the sum of elongations of the two springs.

So, $x\:=\:x_1\:+\:x_2$.12-Jan-2015 Oscillations  C

Therefore, $\frac{k_1+k_2}{k_2}\:=\:\frac{x}{x_1}$.

From this the elongation of the first spring, $x_1\:=\:\frac{k_2 x}{K_1+k_2}$.

Similarly the elongation of the second spring will be $x_2\:=\:\frac{k_1 x}{K_1+k_2}$.

The displacement of the point $P$ is equal to the displacement of the first spring.

Amplitude is the maximum value of displacement of the block.

If $A$ is the amplitude of the block, then the amplitude of the point $P$ will be $A_1\:=\:\frac{k_2 A}{K_1+k_2}$.