
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
410.1k+ 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 Accountancy: Engaging Questions & Answers for Success

Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

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

What organs are located on the left side of your body class 11 biology CBSE

Soap bubble appears coloured due to the phenomenon class 11 physics CBSE

How is the brain protected from injury and shock class 11 biology CBSE

Define least count of vernier callipers How do you class 11 physics CBSE

What is Environment class 11 chemistry CBSE
