AP EAPCET 2025 - 21st May Morning Shift | Simple Harmonic Motion Question 18 | Physics

If the displacement $y$ (in cm ) of a particle executing simple harmonic motion is given by the equation $y=5 \sin (3 \pi t)+5 \sqrt{3} \cos (3 \pi t)$, then the amplitude of the particle is

5 cm

$5(1+\sqrt{3}) \mathrm{cm}$

$5 \sqrt{3} \mathrm{~cm}$

10 cm

AP EAPCET 2025 - 21st May Morning Shift

MCQ (Single Correct Answer)

-0

The angular frequency of a block of mass 0.1 kg oscillating with the help of a spring of force constant $2.5 \mathrm{~N}-\mathrm{m}^{-1}$ is

$02 \mathrm{rad} \mathrm{s}^{-1}$

$5 \mathrm{rad} \mathrm{s}^{-1}$

$10 \mathrm{rad} \mathrm{s}^{-1}$

$2 \mathrm{rad} \mathrm{s}^{-1}$

AP EAPCET 2024 - 23th May Morning Shift

MCQ (Single Correct Answer)

-0

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

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

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

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

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

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