Two simple pendulum first of bob mass M_1 and length L_1 second of bob mass M_2 and length L_2. M_1 = M_2 and L

Q: Two simple pendulum first of bob mass $M_1$ and length $L_1$ second of bob mass $M_2$ and length $L_2$. $M_1 = M_2$ and $L_1 = 2L_2$. If these vibrational energy of both is same. Then which is correct

b (b) $n = \frac{1}{{2\pi }}\sqrt {\frac{g}{l}} $==> $n \propto \frac{1}{{\sqrt l }}$==> $\frac{{{n_1}}}{{{n_2}}} = \sqrt {\frac{{{l_2}}}{{{l_1}}}} = \sqrt {\frac{{{L_2}}}{{2{L_2}}}} $ ==> $\frac{{{n_1}}}{{{n_2}}} = \frac{1}{{\sqrt 2 }}$==> ${n_2} = \sqrt 2 \,{n_1}$ ==> ${n_2} > {n_1}$ Energy $E = \frac{1}{2}m{\omega ^2}{a^2} = 2{\pi ^2}m{n^2}{a^2}$ ==>$\frac{{a_1^2}}{{a_2^2}} = \frac{{{m_2}n_2^2}}{{{m_1}n_1^2}}$ ( $E$ is same) Given ${n_2} > {n_1}$ and ${m_1} = {m_2}$ ==> ${a_1} > {a_2}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

Two simple pendulum first of bob mass $M_1$ and length $L_1$ second of bob mass $M_2$ and length $L_2$. $M_1 = M_2$ and $L_1 = 2L_2$. If these vibrational energy of both is same. Then which is correct

AAmplitude of $B$ greater than $A$
BAmplitude of $B$ smaller than $A$
C
Amplitudes will be same
D
None of these

Medium

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

Download our app
and get started for free

Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*

Download App

Download our appand get started for free

Similar Questions

Download our app
and get started for free