# CBSE Class 9 Physics Revision Worksheet Set A

Read and download free pdf of CBSE Class 9 Physics Revision Worksheet Set A. Students and teachers of Class 9 Physics can get free printable Worksheets for Class 9 Physics in PDF format prepared as per the latest syllabus and examination pattern in your schools. Standard 9 students should practice questions and answers given here for Physics in Grade 9 which will help them to improve your knowledge of all important chapters and its topics. Students should also download free pdf of Class 9 Physics Worksheets prepared by school teachers as per the latest NCERT, CBSE, KVS books and syllabus issued this academic year and solve important problems provided here with solutions on daily basis to get more score in school exams and tests

## Physics Worksheet for Class 9

Class 9 Physics students should refer to the following printable worksheet in Pdf in standard 9. This test paper with questions and answers for Grade 9 Physics will be very useful for exams and help you to score good marks

### Class 9 Physics Worksheet Pdf

CBSE Class 9 Physics Worksheet - Revision

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SOUND AND WAVES

CHARACTERISTICS OF SOUND WAVES:

Sound waves have 4 characteristics
1. Amplitude 2. Wavelength 3. Frequency 4. Speed

• As the sound wave propagates in a medium, the density as well as the pressure of the medium at a given time varies with distance above and below the average value.
• Increase in density is not the same throughout compression. Maximum increase in density is seen at the centre of compression.

• i) Amplitude- The amplitude of sound wave is the height of the crest or tough. The amplitude is how high the crests are. In a sound wave, the maximum displacement associated with the particle constituting a wave is called its amplitude. It is represented by “A’. SI unit is metre.

Amplitude depends upon the force with which an object vibrates . E.g., When we hit a table hard, a loud sound is produced due to its larger amplitude.
Similarly, if we hit the table slowly, the sound produced is low as its amplitude is small. Thus, loudness as well as soft sound is determined by its amplitude.

ii) Wavelength-

The wavelength is the distance between 2 consecutive compressions or 2 consecutive rarefaction is called wavelength n is represented as λ (lambda). Its SI unit is metre. Wavelength can also be considered as the distance over which graph/wave is repeated.

• iii) Frequency- The number of vibrations completed by a particle in one second is the frequency of the sound wave.
Frequency = Number of Oscillations / Total Time. =1/T

We can calculate the frequency of sound by calculating the number of compressions or rarefaction in one second. It is represented by a Greek letter (Greek letter nu). SI unit is Hertz. SI unit of frequency is named after Heinrich Rudolph Hertz who laid foundation 4 future development of radio, telephone, telegraph and TV.

• iv) Time period- The time taken by the particle of the medium for completing one oscillation/vibration is called the time period. It is represented by the symbol “T”. SI unit is second. Time period of a sound wave is the time between 2 successive compressions or 2 successive rarefactions

v) Velocity of sound wave- It is the distance travelled by a wave in one second. Speed is represented by V. Speed with which compression and rarefactions move ahead is called velocity. SI unit is meter per second (m/s).

The speed of sound is more in solids, less in liquids and least in gases

RELATIONSHIP BETWEEN SPEED V, FREQUENCY AND WAVELENGTH OF SOUND

Wave Velocity= Distance covered/ time taken = Wavelength/time taken
= λ /T......................(1)
As =1/T , eq(1), connecting V and in terms of frequency can be written as
V= λ×f......................................(2)
(Or)
Wave velocity= wavelength × Frequency

The velocity of sound remains almost same for all frequencies in a given medium under the same physical conditions.

Questions based on above topics:

Question. What is the formula of time period?
Solution: T= 1/f. Here f stands for frequency.

Question. A wave covers 25 oscillation with its crest and through from point A to B what is the time period of the wave from A to B?
Solution: As we know T= 1/f
Given f = 25 Hz.
( no. Of oscillation of wave = frequency and the SI unit of f is Hz)
T= 1/25 = 0.4Hz.

Question. What affects amplitude of a wave?
Solution: The amount of energy carried by a wave is related to the amplitude of the wave. A high energy wave is characterized by a high amplitude; a low energy wave is characterized by a low amplitude. The energy imparted to a pulse will only affect the amplitude of that pulse.

Question. Does amplitude decrease with distance?
Solution: The energy spreads out in a spherical shell, the energy density decreases as the square of the distance from the source. That's your amplitude decrease. However, the frequency and wavelength stay the same as long as it keeps traveling at the speed of light.

Question. Compute the amplitude of the wave if a wave travels a distance of 0.5 m and has a frequency of 5 Hz..
Solution: Given: Distance D = 0.5 m,
Frequency f = 5 Hz
The amplitude is given by
A= Distance /Frequency
A= 0.5/5 = 0.1 meter.

Question. How are the wavelength and frequency of a sound wave related to its speed?
Solution: Wavelength, speed, and frequency are related in the following way:
Speed = Wavelength x Frequency
v ( speed) = λ × f

Question. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Solution: Frequency = (Number of oscillations) / Total time
Number of oscillations = Frequency × Total time
Given, Frequency of sound = 100 Hz
Total time = 1 min (1 min = 60 s)
Number of oscillations or vibrations = 100 × 60 = 6000
The source vibrates 6000 times in a minute and produces a frequency of 100 Hz.

Question. Calculate the wavelength of a wave whose frequency is 220hz and speed is 440m/sec in medium?
Solution: frequency=velocity/wavelength
So, 220=440/wavelength
wavelength =2m.

Question. Frequency of sound is 100Hz. How many times does it vibrates in a minute?
Solution: Given f= 100 Hz
Time = 1 min
As we know f=1/T
1min = 60 second
F= 1/60
vibration in one minute =100 x 60=6000 times.

Question. A person is listening to a tone of 500Hz sitting at a distance of 450m from the source of sound. What is the time interval between successive compressions from the source?
Solution: Given f= 500 Hz.
T=1/f T=1/500
T=0.002 sec.

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