
The length of sonometer wire AB is $100cm$, where should the two bridges be placed from A to divide the wire in 3 segments whose fundamental frequencies are in the ratio of $1:2:6$
A. $30cm,\,90cm$
B. $60cm,\,90cm$
C. $40cm,\,80cm$
D. $20cm,\,30cm$
Answer
526.5k+ views
Hint: Sonometer is basically used to study the relationship between frequency, tension, and linear mass density and length of a stretched string. A Sonometer is a device based on the principle of Resonance. It is used to verify the laws of vibration of stretched string and also to determine the frequency of a tuning fork.
Complete answer:
Given,
The length of sonometer wire, \[AB = 100cm\]
Ratio of segments \[ = 1:2:6\]
Let ${L_1},{L_2}\,and\,{L_3}$be the lengths of three segments.
Then
$ {L_1} + {L_2} + {L_3} = 100$ ……………...(1)
Also the laws of vibrations of stretched strings
\[{f_1}{L_1} = {f_2}{L_2} = {f_3}{L_3}\]
Given
$ \Rightarrow {f_1}:{f_2}:{f_3} = 1:2:6$
Therefore
$ = {L_1}:2{L_2}:6{L_3}$
So
$ {L_2} = \dfrac{{{L_1}}}{2}$
$ {L_3} = \dfrac{{{L_1}}}{6}$
Put the value in equation (1) and we get
$ \Rightarrow {L_1} + \dfrac{{{L_1}}}{2} + \dfrac{{{L_1}}}{6} = 100$
Simplify
$ \Rightarrow \dfrac{{6{L_1} + 3{L_1} + {L_1}}}{6} = 100$
\[ \Rightarrow 10{L_1} = 100 \times 6\]
\[ \Rightarrow {L_1} = \dfrac{{600}}{{10}}\]
\[ \Rightarrow {L_1} = 60cm\]
Now
$ \Rightarrow {L_2} = \dfrac{{{L_1}}}{2}$
$ \Rightarrow {L_2} = \dfrac{{60}}{2}$
$ \Rightarrow {L_2} = 30cm$
Again
\[ \Rightarrow {L_3} = \dfrac{{{L_1}}}{6}\]
\[ \Rightarrow {L_3} = \dfrac{{60}}{6}\]
\[\Rightarrow {L_3} = 10cm\]
Now
The first bridge, \[{L_1} = 60cm\]
The second bridge, \[{L_1} + {L_2}\]
\[ = 60 + 30\]
\[ = 90cm\]
The second bridge is, \[{L_1} + {L_2} = 90cm\]
So the answer is (2) $60cm,\,90cm$.
Additional Information:
The monochord was used as a musical teaching tool in the 11th century by Guido of Arezzo (c. 990-1050), the musician who invented the first useful form of musical notation.
Note: This is known as resonance - when one object vibrating at the same natural frequency of a second object forces that second object into vibrational motion. The word resonance comes from Latin and means to "respond" - to sound out together with a loud sound. Resonance describes the phenomenon of increased amplitude that occurs when the frequency of a periodically applied force (or a Fourier component of it) is equal or close to a natural frequency of the system on which it acts.
Complete answer:
Given,
The length of sonometer wire, \[AB = 100cm\]
Ratio of segments \[ = 1:2:6\]
Let ${L_1},{L_2}\,and\,{L_3}$be the lengths of three segments.
Then
$ {L_1} + {L_2} + {L_3} = 100$ ……………...(1)
Also the laws of vibrations of stretched strings
\[{f_1}{L_1} = {f_2}{L_2} = {f_3}{L_3}\]
Given
$ \Rightarrow {f_1}:{f_2}:{f_3} = 1:2:6$
Therefore
$ = {L_1}:2{L_2}:6{L_3}$
So
$ {L_2} = \dfrac{{{L_1}}}{2}$
$ {L_3} = \dfrac{{{L_1}}}{6}$
Put the value in equation (1) and we get
$ \Rightarrow {L_1} + \dfrac{{{L_1}}}{2} + \dfrac{{{L_1}}}{6} = 100$
Simplify
$ \Rightarrow \dfrac{{6{L_1} + 3{L_1} + {L_1}}}{6} = 100$
\[ \Rightarrow 10{L_1} = 100 \times 6\]
\[ \Rightarrow {L_1} = \dfrac{{600}}{{10}}\]
\[ \Rightarrow {L_1} = 60cm\]
Now
$ \Rightarrow {L_2} = \dfrac{{{L_1}}}{2}$
$ \Rightarrow {L_2} = \dfrac{{60}}{2}$
$ \Rightarrow {L_2} = 30cm$
Again
\[ \Rightarrow {L_3} = \dfrac{{{L_1}}}{6}\]
\[ \Rightarrow {L_3} = \dfrac{{60}}{6}\]
\[\Rightarrow {L_3} = 10cm\]
Now
The first bridge, \[{L_1} = 60cm\]
The second bridge, \[{L_1} + {L_2}\]
\[ = 60 + 30\]
\[ = 90cm\]
The second bridge is, \[{L_1} + {L_2} = 90cm\]
So the answer is (2) $60cm,\,90cm$.
Additional Information:
The monochord was used as a musical teaching tool in the 11th century by Guido of Arezzo (c. 990-1050), the musician who invented the first useful form of musical notation.
Note: This is known as resonance - when one object vibrating at the same natural frequency of a second object forces that second object into vibrational motion. The word resonance comes from Latin and means to "respond" - to sound out together with a loud sound. Resonance describes the phenomenon of increased amplitude that occurs when the frequency of a periodically applied force (or a Fourier component of it) is equal or close to a natural frequency of the system on which it acts.
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