AP EAPCET 2024 - 23th May Morning Shift | Simple Harmonic Motion Question 20 | Physics

As shown in the figure, two blocks of masses $m_1$ and $m_2$ are connected to spring of force constant $k$. The blocks are slightly displaced in opposite directions to $x_1, x_2$ distances and released. If the system executes simple harmonic motion, then the frequency of oscillation of the system ( $\omega$ ) is

AP EAPCET 2024 - 23th May Morning Shift Physics - Simple Harmonic Motion Question 20 English

$\left(\frac{1}{m_1}+\frac{1}{m_2}\right) k^2$

$\sqrt{\left(\frac{1}{m_1}+\frac{1}{m_2}\right) k^2}$

$\sqrt{\left(\frac{1}{m_1}+\frac{1}{m_2}\right)}$

$\sqrt{\left(\frac{1}{m_1}+\frac{1}{m_2}\right) k}$

AP EAPCET 2024 - 23th May Morning Shift

MCQ (Single Correct Answer)

-0

A mass $M$, attached to a horizontal spring executes simple harmonic motion with amplitude $A_1$. When mass $M$ passes mean position, then a smaller mass millis attached to it and both of them together executing simple harmonic motion with amplitude $A_2$. Then, value of $\frac{A_1}{A_2}$ is

$\sqrt{\frac{m^2+M^2}{M^2}}$

$\sqrt{\frac{m+M}{M^2}}$

$\sqrt{\frac{m+M}{M}}$

$$ \text { } \frac{m+M}{M} $$

AP EAPCET 2024 - 22th May Evening Shift

MCQ (Single Correct Answer)

-0

The displacement of a particle of mass 2 g executing simple harmonic motion is $x=8 \cos \left(50 t+\frac{\pi}{12}\right) \mathrm{m}$, where $t$ is time in second. The maximum kinetic energy of the particle is

160 J

80 J

40 J

20 J

AP EAPCET 2024 - 22th May Evening Shift

MCQ (Single Correct Answer)

-0

The relation between the force ( $F$ in Newton) acting on a particle executing simple harmonic motion and the displacement of the particle ( $y$ in metre) is $500 F+\pi^2 y=0$. If the mass of the particle is 2 g . The time period of oscillation of the particle is

8 s

6 s

2 s

4 s

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