
Two masses ${M_A}$ and ${M_B}$ are hung from two strings of length ${I_A}$ and ${I_B}$ respectively. They are executing SHM with frequency relation ${f_A} = 2{f_B}$ then relation.
(a) ${I_A} = {I_B}/4$, does not depend on mass
(b) ${I_A} = 4{I_B}$, does not depend on mass
(c) ${I_A} = 2{I_B}$ and ${M_A} = 2{M_B}$
(d) ${I_A} = {I_B}/2$ and ${M_A} = {M_B}/2$
Answer
555k+ views
Hint: We all know that we will equate the frequency relation given above in this question and then we will substitute the formula for the frequency to find the relation between the ratios of time and length.
Complete step by step answer:
We know that from the following condition of the question,
${f_A} = 2{f_B}$
Here ${f_A}$ and ${f_B}$ represents the frequency of the springs, respectively.
We will now substitute the formula for the frequency i.e.$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{l}} $
Here l is the length of the string.
$\begin{array}{l}
\left( {\dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_A}}}} } \right) = 2\left( {\dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_B}}}} } \right)\\
\dfrac{1}{{{l_A}}} = 4\left( {\dfrac{1}{{{l_B}}}} \right)\\
{l_A} = \dfrac{{{l_B}}}{4}
\end{array}$
Hence we can see that the ratios do not depend upon mass.
Therefore, ${l_A} = \dfrac{{{l_B}}}{4}$, does not depend upon mass, and the correct option is (A).
Additional Information:
Simple harmonic motion is a special type of periodic motion. The restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards its equilibrium position. It results in an oscillation that continues indefinitely if uninhibited by friction or any other dissipation of energy. The following physical systems are some examples of simple harmonic oscillators: mass on a spring, uniform circular motion, mass on a simple pendulum.
Note:We can see from the above formula that the period as well as the frequency of a vertical spring, as well as the string, does not depend upon masses; rather, it depends upon acceleration due to gravity g and the length of the string l.
Complete step by step answer:
We know that from the following condition of the question,
${f_A} = 2{f_B}$
Here ${f_A}$ and ${f_B}$ represents the frequency of the springs, respectively.
We will now substitute the formula for the frequency i.e.$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{l}} $
Here l is the length of the string.
$\begin{array}{l}
\left( {\dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_A}}}} } \right) = 2\left( {\dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_B}}}} } \right)\\
\dfrac{1}{{{l_A}}} = 4\left( {\dfrac{1}{{{l_B}}}} \right)\\
{l_A} = \dfrac{{{l_B}}}{4}
\end{array}$
Hence we can see that the ratios do not depend upon mass.
Therefore, ${l_A} = \dfrac{{{l_B}}}{4}$, does not depend upon mass, and the correct option is (A).
Additional Information:
Simple harmonic motion is a special type of periodic motion. The restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards its equilibrium position. It results in an oscillation that continues indefinitely if uninhibited by friction or any other dissipation of energy. The following physical systems are some examples of simple harmonic oscillators: mass on a spring, uniform circular motion, mass on a simple pendulum.
Note:We can see from the above formula that the period as well as the frequency of a vertical spring, as well as the string, does not depend upon masses; rather, it depends upon acceleration due to gravity g and the length of the string l.
Recently Updated Pages
Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

How many 5 digit telephone numbers can be constructed class 11 maths CBSE

Draw a well labelled diagram of reflex arc and explain class 11 biology CBSE

What is the difference between noise and music Can class 11 physics CBSE

Trending doubts
In what year Guru Nanak Dev ji was born A15 April 1469 class 11 social science CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

10 examples of friction in our daily life

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

