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
544.5k+ 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.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

How many of the following diseases can be studied with class 11 biology CBSE

Which of the following enzymes is used for carboxylation class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

