Tamil Nadu Board Solutions for Chapter: Oscillations, Exercise 3: EVALUATION

The displacement of a simple harmonic motion is given by $y\left(t\right)=A \sin (\omega t+\phi )$ where $A$ is amplitude of the oscillation, ($\omega$ is the angular frequency and $\phi$ is the phase. Let the amplitude of the oscillation be $8 \mathrm{cm}$ and the time period of the oscillation is $24 \mathrm{s}$. If the displacement at initial time $(\mathrm{t}=0 \mathrm{s})$ is $4 \mathrm{cm}$, then the displacement at $t=6 \mathrm{s}$ is

A pendulum is hung in a very high building oscillates to and from motion freely like a simple harmonic oscillator. If the acceleration of the bob is $16 m{s}^{-2}$ at a distance $4 m$ from the mean position, then the time period is

A hollow sphere is filled with water. It is hung by a long thread. As the water flows out of a hole at the bottom, the period of oscillation will

The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are 

Let the total energy of a particle executing simple harmonic motion with angular frequency is $1 \mathrm{rad} {\mathrm{s}}^{-1}$ is $0.256 \mathrm{J}$. If the displacement of the particle at time $\mathrm{t}=\frac{\mathrm{\pi }}{2} \mathrm{s}$ is $8\sqrt{2} \mathrm{cm}$ then the amplitude of motion is

A particle executes simple harmonic motion and displacement $y$ at time $t_0, 2t_0$ and $3t_0$ are $A, B$ and $C$, respectively. Then the value of $\left(\frac{A+C}{2B}\right)$ is

A mass of $3 \mathrm{kg}$ is attached at the end of a spring moves with simple harmonic motion on a horizontal frictionless table with time period $2\pi$ and with amplitude of $2 \mathrm{m}$, then the maximum fore exerted on the spring is

Given an one dimensional system with total energy $E=\frac{{p_x}^2}{2m}+V\left(x\right)=$ constant, where $p_x$ is the $x$ component of the linear momentum and $V\left(x\right)$ is the potential energy of the system. Show that total time derivative of energy gives us force $F_x=-\frac{d}{dx}V\left(x\right)$. Verify Hooke’s law by choosing potential energy $V\left(x\right)=\frac{1}{2}k{x}^2$.