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Thank U Darshit Mirani Sir for the Book...
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MELDE’S EXPERIMENT
Aim : To verify the laws of vibrating stretched string by Melde’s experiment.
Apparatus : Melde’s apparatus, Thin string, Pan, Weight box, Meter rule.
Theory : Frequencies corresponding to each normal mode of oscillation of vibrating
string are called eigen or proper frequencies or they are also known as
characteristic or natural frequencies. If the string is set oscillating by some
oscillating system and if the frequency of this oscillating system is the same
as any frequency of normal mode of vibrating string, the string vibrates with
large amplitude. Such position is obtained in sonometer and Melde’s
experiment.
The Melde’s apparatus consists of a thin string tied with one prong of the
tuning fork mounted on a wooden stand. The other end of the string is passed
over a small pulley. A light pan is tied with this end. To apply tension on the
string, necessary weights are put in the pan. The hook on one prong of tuning
fork and the pulley are both kept at the same height.
The tuning fork is made to vibrate. Then transverse waves are produced
which passes on to the string. These waves are reflected from the pulley.
Interference is produced between original waves and reflected waves. As a
result, stationery waves are produced and loops are formed on the string.
The tuning fork is adjusted in turn in two different positions.
A POSITION : When the line joining the prongs of the tuning fork
is along the direction of string AB.
In A position, the frequency of vibrating string is twice the frequency of
tuning fork. Thus, we get one oscillation of string per two oscillations of
tuning fork.
Frequency of vibrating string when one loop is produced on it.
f
If N is frequency of tuning fork,
f
N
N
Where, P = No. of loops
L = Length of string
m = Mass per unit length of string
B POSITION : When the line joining two prongs of the tuning fork
is perpendicular to the string. In B position, the frequency of vibrating
string is equal to the frequency of a tuning fork.
N
The symbols have the usual meanings as above.
LAWS OF VIBRATING STRING.
(1) If tension T applied to the string is kept constant, P/L is constant as N
for tuning fork and m for the string are constant.
(2) When the length of the string L is kept constant, P2T is constant.
(3) For a given tuning fork and string, if the no. of loops is to be kept
constant by adjusting the length and tension, then is constant.
Procedure : FIRST LAW : To verify P/L = constant for T constant.
(1) Adjust the tuning fork in A position.
(2) Put some weight ‘M’ in a pan of weight ‘M’, and apply suitable tension
T = [Mo + M]g (dynes) on the string.
(3) Keep the tuning fork near pulley and vibrate it.
(4) Move the tuning fork away from pulley and adjust the length of string
such that one loop is produced on the string.
(5) Vibrate the tuning fork again at this distance and adjust the length of
the string more accurately so as to get clear distinct loops. Measure this
length.
(6) Repeat for two and three loops.
(7) Now adjusting tuning fork in B position and vibrating it for above
determined average lengths, double number of loops are observed.
(8) Take readings for 2, 4 and 6 loops.
SECOND LAW : To verify P2T is constant if length ‘L’ of string is constant.
(1) Adjust tuning fork in A position.
(2) Keep the distance of tuning fork from the centre of pulley constant.
Therefore L = Constant.
(3) Initially, put small weight in pan and vibrate tuning fork.
(4) If more than three loops are obtained, increase the weight in pan to get
three loops.
(5) Find average weight from three distinct observations for 3 loops.
(6) On increasing the weight, the number of loops decreases.
(7) Repeat the experiment for one and two loops.
(8) Keeping the same length and same weights, the number of loops will
double in B position.
THIRD LAW : To verify = constant if number of loops P, are constant.
(1) Adjust the tuning fork in A position.
(2) Put some weight in pan and vibrate the tuning fork.
(3) Obtain two loops and note the corresponding length of the string and
mass ‘M’.
(4) Find average length L for two loops from three observations.
(5) Find the average lengths of a string for two loops for different weights.
(6) Adjust the tuning fork in B position.
(7) Find the average lengths of the string for two loops for three different
weights.
Observation Table :
[1] FIRST LAW :
(1) Mass of pan Mo = _________ gm
(2) Mass in pan M = __________ gm
(3) Tension in string T = (Mo + M)g = __________ dynes
Position of fork
Number of loops
P
Length of string when loops are formed
P/LMean P/LL1 (cm) L2 (cm) Mean
L (cm)
A
B
(1) Graph : L vs P
(1) Mean P/L from calculation for A Position = _____________
(2) Mean P/L from graph for A Position = _____________
(3) Mean P/L from calculation for B Position = _____________
(4) Mean P/L from graph for B Position = _____________
Calculations : Calculate the frequency of tuning fork N for both A and B positions from
using
For A position,
N
And for B position,
N
[2] SECOND LAW :
(1) Length of the string L = _________ cm
(2) Mass of the pan Mo = __________ g
Position of fork
Weight in Pan Tension T = (Mo + M)g
(dyne)1/T
No. of loops
PP2 P2TM1
(gm)M2
(gm)Mean
M (gm)
A
B
(2) Graph : P2 vs 1/T (dyne)
(1) Mean P2T from calculation for A Position = _____________
(2) Mean P2T from graph for A Position = _____________
(3) Mean P2T from calculation for B Position = _____________
(4) Mean P2T from graph for B Position = _____________
Calculations : Calculate the frequency of tuning fork N for both A and B positions from
using
For A position,
N
And for B position,
N
[3] THIRD LAW :
(1) Number of loops = P = _________
(2) Mass of pan Mo = __________ gm
Position of fork
Weight in Pan in (gm)
Tension T = (Mo + M)g
(dynes)
Length of stringL1
(cm)L2
(cm)L
(cm)
A
B
(3) Graph : L (cm) vs
(1) Mean from calculation for A Position = _____________
(2) Mean from graph for A Position = _____________
(3) Mean from calculation for B Position = _____________
(4) Mean from graph for B Position = _____________
Calculations : Calculate the frequency of tuning fork N for both A and B positions from
using
For A position,
N
And for B position,
N
Diagrams :
Result : (1) The laws of vibrating stretched string are verified.
(2) The frequency of tuning fork obtained for A position is ___________ Hz. and
for B position is _________ Hz.
Viva Questions : (1) Explain the following terms :
(a) node
(b) antinode
(c) longitudinal waves
(d) transverse waves
(2) State the laws of vibrating strings.
(3) What is a sonometer ?
(4) What is the audible range of frequencies ?
(5) What is decibel ?
NEWTON’S RINGS
Aim : To determine the wave length of sodium light using the reflected system of
Newton’s rings.
Apparatus : Newton’s Rings apparatus, Sodium vapour lamp, Vernier calipers,
Microscope and Spherometer.
Theory : The Newton’s ring apparatus consists of an optically plane glass plate P on
which is placed a convex lens L of large focal length. Above the lens, another
glass plate G is arranged at 45º to the horizontal.
Let Dn & Dn+k be the diameter’s of nth and (n + k)th dark rings respectively.
Then,
Where, is the wavelength of light used and R is the radius of curvature of
the lens found out by spherometer.
Procedure : (1) When light from a sodium lamp S is rendered parallel by a short focus
convex lens C, the parallel rays fall on the glass plate G, inclined at 45º to
the horizontal, get reflected and then fall normally on the convex lens
placed over the glass plate G.
(2) The system of bright and dark concentric circular rings are observed
through a microscope, which is arranged vertically above the glass plate G.
(3) The microscope is properly focused so that the rings are seen most clearly.
The rings obtained are shown as in figure (2).
(4) Starting from the centre of the frings system, the microscope is moved
towards the left so that the crosswire (one of them) is tangential to the n th
(20th dark ring).
(5) The microscope reading is taken by working the fine adjustment screw,
moving the microscope to the right.
(6) The cross wire is adjusted to be on 18 th, 16th etc. rings in succession upto
the 2nd ring on left and then readings are taken corresponding to the 2nd, 4th
to 20th dark ring on right as before.
(7) The difference between the reading on the left and right of each ring gives
the diameter D of the respective ring. Hence is calculated.
(8) A graph between D2 and n is drawn and slope of the straight line is
determined.
(9) Wavelength is calculated using the given formula.
Precaution : While taking readings corresponding to each ring, the fine adjustment screw
of the microscope should be worked only in one direction to avoid back lash
error.
Observation Table :
L.C. of Spherometer = ___________ mm.
Average distance between legs of spherometer l = __________ cm.
TABLE – 1
Sr. No.
FOR PLANE GLASS FOR CURVED SURFACE
Pitch Scale
Reading A (mm)
Circular Scale
Reading B (mm)
X = A + (B L. C.) (mm)
Pitch Scale
Reading C (mm)
Circular Scale
Reading D (mm)
Y = C + (D L.C.) (mm)
1
2
3
Mean X = ____________ mm. Mean Y = ____________ mm.
Hence X = ____________ cm. Y = ____________ cm.
Height of lens h = Y – X = ____________ cm.
TABLE – 2
No. of Dark Ring
Microscope Reading Diameter D = (a ~ b)
(mm)D2 (cm2)
(cm2)Left Side
a (mm)Right Side
b (mm)2018
161412108642
k = 10
Mean = ___________ cm2
Diagram :
Calculations : (1) Radius of Curvature
R =
= _____________ cm
(2) Wavelength of sodium light found practically.
=
= _____________ .
(3) Wavelength of sodium light found graphically.
=
= _____________ .
Graph : D2 vs No. of Rings
Result : The wavelength of the sodium light = ____________ practically and
___________ graphically.
Viva Questions : (1) What is interference of light ?
(2) Explain the formation of Newton’s rings.
(3) Explain why thin films are coloured ?
(4) Explain why a soap bubble is coloured in sunlight ?
(5) Explain why in Newton’s rings, circular rings are formed.
(6) Define radius of curvature.
DEFLECTION MAGNETOMETER – I
Aim : To determine the magnetic moment M of the given magnet using the
deflection magnetometer in gauss A and gauss B position separately. Also
determine the pole strength ‘m’ of the magnet.
Apparatus : A bar magnet, deflection magnetometer, scale.
Theory : The magnetic compass is the heart of the deflection magnetometer.
The magnet and pointer are cased in a glass box which is mounted at the
centre of a long narrow, wooden board carrying a scale in cm. The
magnetometer is arranged in either (i) the gauss A or tan A position or (ii) the
gauss B or tan B position.
The gauss A position is when the arms of the magnetometer are set in east
west direction i.e. they are normal to the earth’s horizontal component ‘H’.
The magnet whose magnetic moment is to be found out is set parallel to the
arms of the magnetometer.
The gauss B position is when the arms of the magnetometer are set in north-
south direction i.e., they are parallel to the earth’s horizontal component ‘H’.
The magnet whose magnetic moment is to be found out is set perpendicular
to the arms of the magnetometer.
The magnetic moment is a measure of the strength of the magnet. Its units are
gauss cm3. For a magnet of pole strength ‘m’ and length 2 l the magnetic
moment P = 2ml and points from the north pole to the south pole of the
magnet.
Procedure : (1) Measure the geometric length ‘L’ of the magnet. Hence calculate magnetic
length 2 l.
(2) Arrange the deflection magnetometer in gauss A position.
(3) Keep the magnet on the arms of the deflection magnetometer at a distance
where the deflection is between 30º and 60º.
(4) Note down the distance and the deflections 1, 2.
(5) Reverse the position of the magnet at that distance and record the
deflections 3 and 4.
(6) Repeat the above procedure for the same distance on the opposite arm of
the magnetometer and note down deflections 5, 6, 7 and 8 respectively.
(7) Repeat the above procedure for five such distances.
(8) Arrange the deflection magnetometer in Gauss B position.
(9) Keep the magnet perpendicular to the arms of the magnetometer.
(10) Perform steps 3 – 7.
Observations : Geometric length of the bar magnet L = ___________ cm
Magnetic length of the bar magnet l = ___________ cm
Observation Table :
Position Distance Deflection Mean
tan 1 2 3 4 5 6 7 8
Gauss A
Gauss B
Formulae Used : (1) Geometric length of the magnet = L cm
(2) Magnetic length l = L
(3) According to tangent law, F = H tan where, H = 0.36 gauss
(4) For Gauss A position F =
Magnetic moment M =
(5) For Gauss B position F =
Magnetic moment M = . H tan (gauss cm3)
Magnetic pole strength m = gauss cm2
Calculations :
(1) For Gauss A position,
M = tan
m =
(2) For Gauss B position,
M = H tan
m =
Result : The magnetic moment ‘M’ of the given bar in
Gauss A position = _______________ gauss cm3.
Gauss B position = _______________ gauss cm3.
The Pole strength ‘m’ of the given bar magnet in
Gauss A position = _______________ gauss cm2.
Gauss B position = _______________ gauss cm2.
Viva Questions : (1) What is magnetic moment ? Define. Give its unit.
(2) What is pole strength ? What is its unit ?
(3) What is the tangent law ?
(4) Give units of magnetic field strength in both c.g.s and m.k.s system.
DEFLECTION MAGNETOMETER – II
: NEUTRALIZATION METHOD :
Aim : To compare the magnetic moments of two bar magnets by the method of
neutralization using deflection magnetometer.
Apparatus : Two bar magnets, deflection magnetometer, scale.
Theory : If the deflection is produced by the first magnet is neutralized by the second
magnet, then the magnetic field produced by two magnets are the same at the
centre of the compass box, i.e. B1 = B2.
Procedure : (1) Measure the geometric length L and hence calculate the magnetic length l
of the bar magnets.
(2) Arrange deflection magnetometer in gauss A position.
(3) Keep one magnet on one arm of the magnetometer such that the deflection
is between 30º and 60º. Note the distance d1 of the magnet from the
compass.
(4) Keep the second magnet on the opposite arm such that the deflection is
reduced to zero. Note down the distance.
(5) Repeat the experiment for different values of distances.
(6) Repeat the same procedure by keeping the magnet in gauss B position.
Formulae Used:
GAUSS A POSITION
For B1 = B2
GAUSS A POSITION
,
For B1 = B2
Observation : (1) Geometric length of 1st magnet L1 = ____________ cm.
(2) Geometric length of 2nd magnet L2 = ____________ cm.
(3) Magnetic length of 1st magnet l1 = ____________ cm.
(4) Magnetic length of 2nd magnet l2 = ____________ cm.
Observation Table :
Position of Magnetometer
Distance of 1st
Magnet d1 (cm)
Distance of 2nd
Magnet when deflection is zero d2 (cm)
Mean
Gauss A
Gauss B
Calculations :
GAUSS A POSITION
GAUSS A POSITION
Result : The ratio of magnetic moments of given bar magnets in gauss A position is
___________ and in gauss B position is ___________.
Viva Questions : (1) Define magnetic moment.
(2) What is magnetic length ?
(3) What is the difference between geometric and magnetic lengths of a
bar magnet ?
(4) What is null point ?
(5) Explain why null points are found on the equatorial line of a magnet
when it is placed with its north pole pointing north.
(6) Explain why null points are found on the axial line of a magnet when
it is placed with its south pole pointing north.