
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
560.4k+ 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.
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