Experiment 6 : To find the frequency of the A.C. mains with a sonometer (and an electromagnet).
Aim :
To find the frequency of the A.C. mains with a sonometer (and an electromagnet).
Apparatus :
A sonometer with a soft iron wire stretched over it, an electromagnet, a step down transformer, slotted half kilogram weights, a hanger, physical balance and a weight box.
Theory :
When a wire of length l having mass m per unit length under tension T is made to vibrate, its fundamental frequency is given by :
$n = \frac{1}{2l} \sqrt{\frac{T}{m}}$
When AC is passed in the coil of electromagnet, it is magnetised twice in every cycle, first with one of its face as N-pole and then with same face as S-pole. If the electromagnet is held close to the middle of the sonometer wire (Fig. 4.4), the wire will be attracted twice during each cycle towards the electromagnet and forced vibrations are produced. If the length of the sonometer wire between two wedges is so adjusted that it is set into resonant vibrations by AC, then the frequency of AC will be half that of the frequency of the wire. If f is the frequency of A.C., then
$f = \frac{n}{2} = \frac{1}{4l} \sqrt{\frac{T}{m}}$
Procedure :
- Set up the apparatus as shown in Fig. 4.4. Stretch the steel wire PQ of sonometer by placing a load of 1kg in the hanger.
- Support the electromagnet in a stand such that its one pole lies close to the middle of the sonometer wire.
- Connect the electromagnet to the secondary of a step down transformer.
- Start the current through the electro-magnet and adjust the position of wedges until the wire start vibrating.
- Adjust the length of the wire between the wedges by sliding one of them keeping the rider always in the middle till it flies off.
- Measure the vibrating length and note the tension in the string.
- Repeat four times the above process by increasing the load in steps of $\frac{1}{2} kg$.
- Switch off AC supply and calculate mass of 100 cm sonometer wire. Hence find mass per unit length m for the wire.
Observations :
Length of wire l = 140 cm = 1.4 m
Mass of wire M = 6.6 g = 0.0066 Kg
Mass per unit length $m = \frac{M}{l} =0.0047 kg m^{-1}$
Acceleration due to gravity g = $9.8 m s^{-2}$.
Calculations :
Frequency of AC mains, $f = \frac{n}{2}$ = 49.2 Hz
Actual frequency of AC supply = 50 Hz
Difference = 50-49.2 = 0.8 Hz
$\% \text{ Error} = \frac{\text{Difference}}{\text{Actual Value}} \times 100$
$\% \text{ Error} = \frac{50 - 49.2}{50} \times 100$
$\% \text{ Error} = \frac{0.8}{50} \times 100 = 1.6\%$
Precautions :
- Sufficient load should be put on the wire to make it light.
- The sonometer wire should be free from kinks and stretched horizontally.
- Pulley should be frictionless.
- For each load, the length should be taken at least twice.
- The pole of the electromagnet should be placed very close to the middle of vibrating segment of string.
Sources of Error :
- AC frequency may not be stable.
- Sonometer wire may not be uniform.
- Pulley may have some friction.
VIVA VOCE :
Q. 1. What is meant by frequency of A.C.?
Ans. Alternating current varies with time. It completes a cycle when it varies from zero to maximum, max. to zero, zero to maximum in reverse direction and from this maximum to zero. "The number of cycles per second is called frequency of A.C."
Q. 2. What is the frequency of the mains of your city supply?
Ans. 50 cycles/sec. (or Hz).
Q. 3. How many times current becomes zero in one second?
Ans. 100 times.
Q. 4. What is the e.m.f. of the A.C. mains?
Ans. 220V.
Q. 5. Is this the peak value?
Ans. No, it is r.m.s. value.
Q. 6. What is the peak value?
Ans. $220\sqrt{2}$V.
Q. 7. What is the function of holes in the sonometer box?
Ans. To make the inside air in communication with external air.
Q. 8. What are free vibrations?
Ans. Vibrations of a body is said to be free if it vibrates with its own natural frequency without the help of any external periodic force.
Q. 9. What are forced vibrations?
Ans. If a body vibrates under the influence of the other vibrating body then the vibrations of the first body are said to be forced vibrations. The frequency of the forced vibrations is equal to the frequency of the driver (which is the tuning fork in this case).
Q. 10. What is resonance?
Ans. When a body is made to perform forced vibrations under the influence of another vibrations body and if the frequency of the forced vibrations becomes equal to that of the natural frequency of the body forced to vibrate, then the vibrations are said to be resonant vibrations. At this instant the body starts vibrating with a large amplitude.
Q. 11. What is the natural frequency of a sonometer wire under tension?
Ans. The natural frequency 'n' of the sonometer wire vibrating in one loop is given by:
$n = \frac{1}{2l} \sqrt{\frac{T}{m}}$
where T is the tension, m is the mass per unit length of the wire and l is the length of the wire between the bridges.
Q. 12. Why does the string of the sonometer vibrate and resonant vibrations are produced?
Ans. When the electromagnet is held close to the middle of the sonometer wire between the wedges, the wire is attracted twice during each cycle towards the electromagnet. Forced vibrations are produced in the wire. The length of sonometer wire between the wedges is adjusted to get resonance condition.
Q. 13. How do you detect that the condition of resonance has reached?
Ans. A small piece of paper known as rider (V-shaped paper) is kept near the middle of the wire. The length of vibrating wire between the bridges is adjusted with their help of the bridges so that the rider flies of when the vibrating tuning fork is placed on the board.
Q. 14. How do you ensure that the wire between the two bridges is vibrating in a single loop?
Ans. In order to ensure this, the resonating length of the wire should be attained by increasing the distance between the bridges starting initially with minimum possible.
Q. 15. How is the energy communicated from wire to the sound box?
Ans. Through the wedges.
Q. 16. What is the role of step down transformer?
Ans. It is used to decrease alternating voltage.
Q. 17. Can we use brass wire in place of iron wire of the sonometer?
Ans. No, the wire must be some magnetic material, otherwise it will not be attracted by electromagnet and vibrations are not produced.
Q. 18. Define wave.
Ans. When a medium is disturbed, the pattern of disturbance which moves without the actual physical transfer of matter of the medium as a whole is called wave.
Q. 19. Define longitudinal waves.
Ans. Those waves in which the particles of the medium vibrate to and fro about their mean position along the direction of wave are called longitudinal waves.
Q. 20. Define transverse waves.
Ans. Those waves in which the particles of the medium vibrate up and down perpendicular to the direction of propagation of the wave are called transverse wave.
Q. 21. What types of waves are produced in sonometer wire?
Ans. Transverse stationary waves.
Q. 22. Why do we say that sound waves are mechanical in character?
Ans. Because sound waves can travel with the help of mechanical oscillations of the particles of the medium.
Q. 23. What is the range of audible sound waves?
Ans. Audible sound waves have frequency between 20 Hz to 20 kHz.
Q. 24. Why the instrument is called a sonometer?
Ans. Because it measures the frequency of sound. Sono means sound.
Q. 25. Does the transformer affect the frequency of a.c.?
Ans. No.
