# Oscillations - Class 11 - Physics

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## Extra Questions - Oscillations | NCERT | Physics | Class 11

A particle in SHM is described by the displacement equation $x(t) = A \cos (\omega t + \theta)$. If the initial $(t = 0)$ position of the particle is $1 \mathrm{~cm}$ and its initial velocity is $\pi \mathrm{cm/s}$, what is its amplitude? The angular frequency of the particle is $\pi \mathrm{s}^{-1}$.

A) $1 \mathrm{~cm}$

B) $\sqrt{2} \mathrm{~cm}$

C) $2 \mathrm{~cm}$

D) $2.5 \mathrm{~cm}$

To solve for the amplitude of a simple harmonic motion (SHM), we begin by utilizing the equation for displacement:

$$ x(t) = A \cos (\omega t + \theta) $$

At time $t = 0$, the particle's position $x(0) = A \cos(\theta)$ is given as $1 \text{ cm}$. The velocity function in SHM is derived through differentiation of the displacement:

$$ v(t) = -A\omega \sin(\omega t + \theta) $$

Again, when $t = 0$, this simplifies to $v(0) = -A \omega \sin(\theta)$. Given that the initial velocity $v(0) = \pi \text{ cm/s}$ and the angular frequency $\omega = \pi \text{ s}^{-1}$, combining these values we get:

$$ \pi = -A \pi \sin(\theta) $$

Thus,

$$ \sin(\theta) = -\frac{1}{A} $$

Combining our knowledge, $x(0) = A \cos(\theta) = 1 \text{ cm}$, and we know $A\cos(\theta)^2 + A\sin(\theta)^2 = A^2$ (from trigonometric identity $\cos^2(\theta) + \sin^2(\theta) = 1$), solving for A:

$$ 1^2 + \left(-\frac{\pi}{\pi A}\right)^2 = A^2 $$

$$ 1 + \frac{1}{A^2} = A^2 $$

This simplifies to:

$$ A^4 - A^2 - 1 = 0 $$

Let $u = A^2$, so:

$$ u^2 - u - 1 = 0 $$

This equation can be solved using the quadratic formula:

$$ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} $$

Choosing the positive root for $u$ because amplitude is always positive:

$$ A^2 = \frac{1 + \sqrt{5}}{2} $$

Therefore, the amplitude $A$ is:

$$ A = \sqrt{\frac{1 + \sqrt{5}}{2}} $$

But initially we are trying to find a straight computation mistake in the setup:

Our initial setup was for simplicity led us astray. Actually, back to our initial derived relationship with proper values:

$$ \pi = \pi \sqrt{A^2 - 1} $$

Substituting $\omega$:

$$ \pi = \pi \sqrt{A^2 - 1} $$

Divide both sides by $\pi$:

$$ 1 = \sqrt{A^2 - 1} $$

Square both sides to remove the square root:

$$ 1 = A^2 - 1 $$

$$ A^2 = 2 $$

So,

$$ A = \sqrt{2} \ \text{cm} $$

Hence, the correct option is **$\sqrt{2} \ \text{cm}$** (Option B).

A particle starts executing S.H.M. of amplitude a and total energy E. At the instant its kinetic energy is $\frac{3 E}{4}$, its displacement y is given by

A) $y=\frac{1}{\sqrt{2}}$

B) $y=\frac{1}{2}$

C) $y=\frac{a \sqrt{3}}{2}$

D) $y=a$

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A smooth piston of mass $m$ and cross-section area $A$ is in equilibrium on a diatomic gas filled in a cylindrical container and there is vacuum above the piston. The piston and container are non-conducting, and the height of the gas column in the container is $l_{0}$ at equilibrium. If the piston is displaced slightly down from the equilibrium position, then the time period of its oscillation will be:

A) $2\pi\sqrt{\frac{10 l_{0}}{7 g}}$

B) $2\pi\sqrt{\frac{7 l_{0}}{10 g}}$

C) $2\pi\sqrt{\frac{7 l_{0}}{5 g}}$

D) $2\pi\sqrt{\frac{5 l_{0}}{7 g}}$

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A simple pendulum is made of a body which is a hollow sphere containing mercury, suspended by means of a wire. If a little mercury is drained off, then the time period of the pendulum:

A) Decreases B) Increases

C) Remains unchanged

D) Becomes erratic