What are the Characteristics of Light

What are the Characteristics of Light

The Characteristics of light :     

  1. Light is an electromagnetic wave.
  2. Light travels in a straight line.
  3. Light is a transverse wave, and does not need any medium to travel. Light can travel through vaccum. Its speed through vaccum is 3 × 108 m/s.
  4. The velocity of light changes when it travels from one medium to another.
  5. The wavelength (λ) of light changes when it goes from one medium to another.
  6. The frequency (f) of the light wave remains the same in all media.
  7. Light gets reflected back from polished surfaces, such as mirrors, polished metal surfaces, etc.
  8. Light undergoes refraction (bending) when it travels from one transparent medium to another.
  9. Light does not need a material medium to travel, that is, it can travel through a vacuum too. Scientists have assigned a value of 299, 792, 458 m/s to the speed of light in vacuum.
  10. According to current scientific theories, no material particle can travel at a speed greater than that of light in vacuum.

What is Collection and Organisation of Data

What is Collection and Organisation of Data

Data

Mr Gupta has done a survey in a reputed school to find out the favourite game of the students.
The information collected from the students of Class VI to Class XII are as follows:

GameBasket ballVolley ballFoot ballCricketTable tennisBadminton
Number of students205215600560156264

From the above table, we can easily visualise the following information:
(a) Number of students present on the day of survey.
(b) Which game has got the maximum popularity?
(c) Which game has got the minimum popularity?
(d) What is the fraction of the number of students who like badminton?
The initial step of any investigation is the collection of data. It may be a collection of numbers, figures, facts, or symbols. The information needed or the data collected must be relevant to the need. Collection of information in the form of numerical figures is called data. Literal meaning of data is known facts or facts collected to draw some inference.
There are two types of data—primary data and secondary data.

Primary Data
The data collected directly from the source is called the primary data. For example, the data collected by Mr. Gupta is primary data.
Secondary Data
When the data is collected from an external source, it is called the secondary data. For example,
the data collected from newspapers, magazines, internet, etc. is secondary data.

Note:
Data can be classified as primary and secondary data.
The data collected directly for the first time by the observer is called the primary data.
The data collected from any external source like TV, internet, newspaper, etc. is called the secondary data.

Use of Data Collection

In order to preserve data, we need to collect the data, organise them, and record them so that whenever we want to use the information, we can refer to it.
For example,
(a) Taking attendance by the teacher in the class
(b) Different types of games played by the children
(c) Popular brand of cars

Organisation of data

Organisation of data helps in bringing about meaningful conclusion from the data. For example, a group of students of Class VI took part in a music competition. The judges gave the following grades: 2, 4, 5, 1, 3, 2 etc. Does it make any sense to you? This is data but not meaningful. To make the data meaningful, we have to arrange the data in a tabular form.

Name of the studentGrade
Raju2
Lata4
Seema5
Rehana1
Ajay3
Rekha2
Arnav2
Peter4
Tripti5

Hence, understanding and analysing the data become easier by means of tables or pictures.

Frequency
Let us consider an example:
The coach of a cricket team wants to know the runs scored by the players of both the teams in the previous match. He will call each and every player and note down their runs as given below:
20, 35, 40, 25, 47, 50, 20, 35, 86, 20, 38, 75, 98, 20, 40, 50, 102, 20, 38, 40, 86, 75, 25, 38
This way of recording the data does not help him to answer the following questions:
(a) What is the maximum runs scored?
(b) How many players made a half-century?
(c) How many players scored 75 runs or more than 75 runs?
(d) Difference between the maximum and minimum runs scored.
Now, we write the above data in ascending or descending order:
20, 20, 20, 20, 20, 25, 25, 35, 35, 38, 38, 38, 40, 40, 40,47, 50, 50, 75, 75, 86, 86, 98,102
This is a better way of writing the collected data but it is time-consuming and there is always a possibility of missing some numbers, if the information is collected for a larger number of persons. It may he helpful in answering the 215 questions as given in (a) and (d), but we might find it difficult to answer the questions as given in (b) and (c).

When the number of observation is larger, to minimise the number of errors and to make tabulation easier, we can use tally marks. In the tally marks method we draw three columns. In the first column, write runs, in the second column, tally marks, and in the third column, the number of players. Tally marks are always recorded in bunches of five. Fifth tally mark is drawn diagonally across the first four to make a group of five.

collection-organisation-data

Finally after counting tally marks the actual numbers can be written in the third column which represent frequency. Now, looking at the table we can easily answer the questions (b) and (c), i.e., 2 and 6 respectively.

Note:

  • The singular form of data is datum. Data is in plural form but now a days, data is used in both singular and plural forms.
  • A tabular form is a form where you can find all information and data in several rows and columns.
  • Frequency is the number of times of a repeating event.
  • Tally is a symbol like a vertical bar as shown in the chart against each data or item.

What Is The Relation Between Wave Velocity, Frequency And Wavelength

Relation Between Wave Velocity, Frequency And Wavelength For A Periodic Wave

\( \text{wave velocity}=\frac{\text{distance covered}}{\text{Time}\,\,\text{taken}} \)
\( =\frac{\text{wave}\,\text{length}}{\text{Time}\,\text{taken}} \)
\( \text{or v}=\frac{\lambda }{T}\text{ }……\text{ (1)} \)
\( \text{since }\!\!~\!\!\text{ }v=\frac{\text{1}}{\text{T}}\text{, equation }\left( \text{1} \right)\text{ can also be written as} \)
\( \text{v}=v\lambda \text{ }……\text{ (2)} \)
Wave Velocity = Frequency × Wave Length

Relation Between Wave Velocity, Frequency And Wavelength Example Problems With Solutions

Example 1:    If 50 waves are produced in 2 seconds, what is its frequency ?
Solution:    Frequency,
\( v=\frac{\text{Number}\,\text{of}\,\text{wave}\,\text{produced}}{\text{Time}\,\text{taken}} \)
\( =\frac{50}{2}=25\text{ Hz} \)

Example 2:    A source produce 50 crests and 50 troughs in 0.5 second. Find the frequency.
Solution:    1 crest and 1 trough = 1 wave
∴ 50 crests and 50 troughs = 50 waves
\( \text{Now, Frequency, }v=~\frac{\text{Number}\,\text{of}\,\text{wave}}{\text{Time}} \)
\( =\frac{50}{0.5}=100\text{ Hz} \)

Example 3:    Sound waves travel with a speed of 330 m/s. What is the wavelength of sound waves whose frequency is 550 Hz ?
Solution:    Given velocity, v = 330 m/s,
Frequency, \(\upsilon\) = 550 Hz
\(\therefore \text{ }wavelength,~~\text{ }\lambda =\frac{\text{v}}{\upsilon }\)
\( =\frac{330}{550}=0.6\text{ }m \)

Example 4:    The wave length of sound emitted by a source is 1.7 × 10-2 m. Calculate frequency of the sound, if its velocity is 343.4 ms-1.
Solution:    The relation ship between velocity, frequency and wave length of a wave is given by the formula v = \(\upsilon\) × λ
Here, velocity, v = 343.4 ms-1
frequency \(\upsilon\) = ?
and wavelength, λ= 1.7 × 10-2 m
So, putting these values in the above formula, we get :
343.4 = \(\upsilon\) × 1.7 × 10-2
\( v=\frac{343.4}{1.7\times {{10}^{-2}}} \)
\( =\frac{343.4\,\times {{10}^{2}}}{1.7} \)
= 2.02 × 104 Hz
Thus, the frequency of sound is 2.02 × 104 hertz.

Example 5:    A wave pulse on a string moves a distance of 8m in 0.05 s.
(i) Calculate the velocity of the pulse.
(ii) What would be the wavelength of the wave on the same string, if its frequency is 200 Hz ?
Solution:    (i) Velocity of the wave,
\( \text{v}=\frac{\text{Distance covered}}{\text{Time}\,\text{taken}}=\frac{8m}{0.05s}=160\text{ m/s} \)
(ii) Periodic wave has the same velocity as that of the wave pulse on the same string.
\( \therefore \text{ Wavelength, }\lambda =~\frac{\text{v}}{v}=~\frac{160\,m/s}{200\,Hz}~=0.8\text{m} \)
Thus, the wavelength of the wave is 0.8 m.

Example 6:    A person has a hearing range of 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies ? Take the speed of sound in air as 340 m/s.
Solution:    Given : \({{v}_{1}}\) = 20 Hz, V = 340 m/s
\( \therefore \text{ }{{\lambda }_{1}}=\frac{\text{v}}{{{v}_{1}}}=\frac{340}{20}=17\text{ m} \)
\({{v}_{2}}\) = 20 kHz = 20,000 Hz, v = 340 m/s
\( \therefore \text{ }{{\lambda }_{2}}=\frac{\text{v}}{{{v}_{2}}}=\frac{340}{20,000}=1.7\times {{10}^{-2}}m=1.7cm \)
∴ The typical wavelengths are 17 m and 1.7 cm.

Example 7:    A longitudinal wave is produced on a toy string. The wave travels at a speed of 30 cm/s and the frequency of the wave is 20 Hz. What is the minimum separation between the consecutive compressions of the string ?
Sol. Given, Velocity, v = 30 cm/s
Frequency, \(\upsilon\) = 20 Hz
Minimum separation between the two consecutive compressions is equal to one wavelength λ and
\(\lambda =\frac{\text{v}}{v}=\frac{30\,cm/s}{20\,Hz}=1.5\text{ cm}\)

Example 8:    Wave of frequency 200 Hz produced in a string is represented in figure. Find out the following
Wave-of-frequency
(i) amplitude
(ii) wavelength
(iii) wave velocity
Solution:    (i) Amplitude = Maximum displacement = 10 cm
(ii) Wavelength λ = Distance between two successive crests = 40 cm
(iii) Now, frequency, n = 2 Hz
Wavelength, λ= 40cm = 0.4 m
∴ Wave velocity, v = \(\upsilon\)λ
= 200 × 0.4 m/s
= 80m/s

Example 9:    A stone is dropped into a well 44.1 m deep. The sound of splash is heard 3.13 seconds after the stone is dropped. Calculate the velocity of sound in air.
Solution:    First we calculate the time taken by the stone to reach the water level by using the relation:
s = ut + \(\frac { 1 }{ 2 }\) gt2
Here s = 44.1 m, u = 0, g = 9.8 m/s2
∴ 44.1 = 0 × t + \(\frac { 1 }{ 2 }\) × 9.8 × t2
\( {{t}^{2}}=\frac{44.1\times 2}{9.8}=9 \)
or t = 3 s
Time taken by the sound to reach the top of the well
t2 = 3.13 – 3 = 0.13 s
Now, speed of sound
\( \frac{\text{Distance}}{\text{Time}}=\frac{44.1\,m}{0.13\,s}=339.2\text{ m/s} \)

What Is Circular Motion

Circular Motion

When a body moves in such a way that its distance from a fixed point always remains constant, then its motion is said to be the circular motion.

Uniform circular motion:
If the radius vector sweeps out equal angles in equal times, then its motion is said to be uniform circular motion.
Uniform-circular-motion
In uniform circular motion speed remains constant.
Linear velocity, being a vector quantity, its direction changes continuously.
The direction of velocity is along the tangent at every point.

Angular velocity:
\(\omega =\frac{\Delta \theta }{\Delta t}\)
A vector quantity
Direction is perpendicular to plane of rotation
Note: If the particle is revolving in the clockwise direction then the direction of angular velocity is perpendicular to the plane downwards. Whereas in case of anticlockwise direction, the direction will be upwards.
Unit is Radian/sec.
In uniform circular motion the direction of angular velocity is along the axis of rotation which is constant throughout.
Angular velocity remains constant in magnitude as well as in direction.
v = rω where r = radius of the circle.

Centripetal acceleration:
In uniform circular motion the particle experiences an acceleration called the centripetal acceleration.
\({{a}_{c}}=\frac{{{v}^{2}}}{r}\)
The direction of centripetal acceleration is along the radius towards the centre.

Centripetal force:
Always acts towards centre.
Centripetal force is required to move a particle in a circle.
Because Fc is always perpendicular to velocity or displacement, hence the work done by this force will always be zero.
Note:
Circular motion in horizontal plane is usually uniform circular motion.
Remember that equations of motion are not applicable for circular motion.

Time period:
It is the time taken to complete one complete revolution.
In one revolution, angle subtended is 2π and if T is time period, then the angular velocity is given by
\(\omega =\frac{2\pi }{T}\text{         or         }T=\frac{2\pi }{\omega }\)

Frequency:
Frequency is defined as the number of revolutions per second.
\(i.e.~\text{    n}=\frac{1}{T}=\frac{\omega }{2\pi }\)

Uniform Circular Motion Example Problems With Solutions

Example 1:    A particle moves in a circle of radius 2 m and completes 5 revolutions in 10 seconds. Calculate the following
(i) Angular velocity and
(ii) Linear velocity.
Solution:    Since, it completes 5 revolutions in 10 seconds.
∴ Time period = 10/5 = 2s
(i) Now angular velocity,
\(\omega =\frac{2\pi }{T}=\frac{2\pi }{2}=\pi \text{ rad/s}\)
(ii) Linear velocity is given by
v = rω = 2π
∴ v = 2π m/s

Example 2:    The length of second’s needle in a watch is 1.2 cm. Calculate the following :
(i) Angular velocity and
(ii) Linear velocity of the tip of the needle.
Solution:    (i) We know that the second’s needle in a watch completes one revolution in 60 seconds.
∴ Time period, T = 60 s
Angular velocity,
\( \omega =\frac{2\pi }{T}=\frac{2\pi }{T}=\frac{\pi }{30}\text{ rad/s}~ \)
(ii) Length of the needle = 1.2 cm = Radius of the circle
Linear velocity of the tip of the needle is given by
v = rω
\( v=1.2\times \frac{\pi }{30}=~\frac{\pi }{25} \)
\( v=\frac{\pi }{2s}=1.266\times {{10}^{-1~}}\text{cm/sec}\text{.} \)

Example 3:    Earth revolves around the sun in 365 days. Calculate its angular velocity.
Solution:    Time period,  T = 365 days
= 365 × 24 × 60 × 60 seconds
∴ Angular velocity
\(\omega =\frac{2\pi }{T}=\frac{2\pi }{365\times 24\times 60\times 60}\)
= 1.99 × 10-7 rad/s.