
The total length of a sonometer wire between fixed ends is 110 cm. Two bridges are placed to divide the length of wire in ratio \[{\mathbf{6}}:{\mathbf{3}}:{\mathbf{2}}\]. The tension in the wire is 400 N and the mass per unit length is \[{\mathbf{0}}.{\mathbf{01}}{\mathbf{kg}}/{\mathbf{m}}\]. What is the minimum common? Frequency with which three parts can vibrate?
A. 1100 H
B. 1000 H
C. 166 H
D. 100 H
Answer
471k+ views
Hint: A sonometer is a device of demonstrating the relation between the frequency of the sound produced by a plucked string, and the tension, length and mass per unit of length of a string and its generalised formula is given by:
$f = \dfrac{l}{{2l}}\sqrt {\dfrac{T}{m}} $
Where,
f = frequency of the sound
l = length of the string
T = tension produced in string
m = mass per unit of length of a string
Complete step by step solution:
Total Length Of The Wire = 110cm = 1.1meter
Since These Parts Are Divided Into 6:3:2ratio
Hence,length of these parts = 0.6:0.3:0.2meter
mass per unit length,( m) = 0.01kg
Tension Produced ( T )= 400N
We know that frequency in a manometer is given by:
$f = \dfrac{l}{{2l}}\sqrt {\dfrac{T}{m}} $
hence,
by using the above formula we can find the frequency of all the three parts and then
a common frequency can be calculated by numerically adding the individual frequency
therefore,
\[{f_1} = \dfrac{1}{{2 \times 0.6}}(\sqrt {\dfrac{{400}}{{0.01}}} )\] ,where $f_1$ means frequency of wave in first part
\[ = \dfrac{{200}}{{2 \times 0.6}}\]
\[ = \dfrac{{1000}}{6}Hz\]
\[{f_2} = \dfrac{1}{{2 \times 0.3}}(\sqrt {\dfrac{{400}}{{0.01}}} )\] ,where $f_2$ means frequency of wave in second part
\[ = \dfrac{{1000}}{3}Hz\]
\[{f_3} = \dfrac{1}{{2 \times 0.2}}(\sqrt {\dfrac{{400}}{{0.01}}} \],where $f_3$ means frequency of wave in third part
\[ = \dfrac{{1000}}{2}\]Hz
Common frequency, i.e.
\[f = {f_1} + {f_2} + {f_3}\]
\[ = \dfrac{{1000}}{6} + \dfrac{{1000}}{3} + \dfrac{{1000}}{2}\]
\[ = 1000Hz\]
It is the frequency with which wave will propagate in all the three parts
Hence, we can see the minimum common frequency in which three parts can vibrate is 1000Hz.
Note: Non-magnetic metallic wire like brass or copper is used as a wire in a sonometer. Frequency is directly proportional to tension (T) produced in the wire and is inversely proportional to the length of the wire (l) and mass per unit length (m) of the wire.
$f = \dfrac{l}{{2l}}\sqrt {\dfrac{T}{m}} $
Where,
f = frequency of the sound
l = length of the string
T = tension produced in string
m = mass per unit of length of a string
Complete step by step solution:
Total Length Of The Wire = 110cm = 1.1meter
Since These Parts Are Divided Into 6:3:2ratio
Hence,length of these parts = 0.6:0.3:0.2meter
mass per unit length,( m) = 0.01kg
Tension Produced ( T )= 400N
We know that frequency in a manometer is given by:
$f = \dfrac{l}{{2l}}\sqrt {\dfrac{T}{m}} $
hence,
by using the above formula we can find the frequency of all the three parts and then
a common frequency can be calculated by numerically adding the individual frequency
therefore,
\[{f_1} = \dfrac{1}{{2 \times 0.6}}(\sqrt {\dfrac{{400}}{{0.01}}} )\] ,where $f_1$ means frequency of wave in first part
\[ = \dfrac{{200}}{{2 \times 0.6}}\]
\[ = \dfrac{{1000}}{6}Hz\]
\[{f_2} = \dfrac{1}{{2 \times 0.3}}(\sqrt {\dfrac{{400}}{{0.01}}} )\] ,where $f_2$ means frequency of wave in second part
\[ = \dfrac{{1000}}{3}Hz\]
\[{f_3} = \dfrac{1}{{2 \times 0.2}}(\sqrt {\dfrac{{400}}{{0.01}}} \],where $f_3$ means frequency of wave in third part
\[ = \dfrac{{1000}}{2}\]Hz
Common frequency, i.e.
\[f = {f_1} + {f_2} + {f_3}\]
\[ = \dfrac{{1000}}{6} + \dfrac{{1000}}{3} + \dfrac{{1000}}{2}\]
\[ = 1000Hz\]
It is the frequency with which wave will propagate in all the three parts
Hence, we can see the minimum common frequency in which three parts can vibrate is 1000Hz.
Note: Non-magnetic metallic wire like brass or copper is used as a wire in a sonometer. Frequency is directly proportional to tension (T) produced in the wire and is inversely proportional to the length of the wire (l) and mass per unit length (m) of the wire.
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