Damped, Forced Oscillations & Superposition

The amplitude of a damped oscillator becomes $\left(\frac{1}{3}\right)^{\text{rd}}$ of original amplitude in 2 seconds. If its amplitude after 6 second become $\left(\frac{1}{n}\right)$ times the original amplitude, the value of $n$ is ( $n$ is non zero integer)

A particle is subjected to two simple harmonic motions as: $x_1 = \sqrt{7}\sin 5t\text{ cm}$ and $x_2 = 2\sqrt{7}\sin\left(5t+\frac{\pi}{3}\right)\text{ cm}$ where $x$ is displacement and $t$ is time in seconds. The maximum acceleration of the particle is $x \times 10^{-2}\text{ m s}^{-2}$. The value of $x$ is:

A particle is performing simple harmonic motion and if the oscillations are damped oscillations then the angular frequency is given by

The two-dimensional motion of a particle, described by $\vec{r} = (\hat{i}+2\hat{j})A\cos\omega t$ is a/an: A. parabolic path

B. elliptical path

C. periodic motion

D. simple harmonic motion

Choose the correct answer from the options given below:

Motion of a particle in x-y plane is described by a set of following equations $x = 4\sin\left(\frac{\pi}{2}-\omega t\right)\text{ m}$ and $y = 4\sin(\omega t)\text{ m}$. The path of the particle will be:

A block of mass 1 kg attached to a spring is made to oscillate with an initial amplitude of 12 cm. After 2 minutes the amplitude decreases to 6 cm. Determine the value of the damping constant for this motion. (take ln 2 = 0.693)

Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass = 500g, Decay constant = 20 g/s then how much time is required for the amplitude of the system to drop to half of its initial value? (ln 2 = 0.693)

The damping force of an oscillator is directly proportional to the velocity. The unit of constant of proportionality is

In damped SHM, the SI unit of damping constant is

The displacement of a damped harmonic oscillator is given by

x(t) = e–0.1t cos (10$\pi$t + f).

Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to:

A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1/1000 of the original amplitude is close to :-

Two simple harmonic motions, as shown below, are at right angles. They are combined to form Lissajous figures.

x(t) = A sin (at + $\delta$)

y(t) = B sin (bt)

Identify the correct match below.

A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 Nm−1 and oscillates in a damping medium of damping constant 10−2 kg s−1. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to:

When two displacements represented by y1 = a sin$(\omega t)$ and y2 = b cos$(\omega t)$ aresuperimposed the motion is

A simple harmonic oscillator of angular frequency 2 rad s${}^{-1}$ is acted upon by an external force F = sint N. If the oscillator is at rest in its equilibrium position at t = 0, its position at later times is proportional to:

x and y displacements of a particle are given as

x(t) = a sin$\omega$t and y(t) = a sin2$\omega$t.

Its trajectory will look like:

A pendulum with time period of 1s is losing energy due to damping. At certain time its energy is 45 J. If after completing 15 oscillations, its energy has become 15 J, its damping constant (in s${}^{-1}$) is:

A particle which is simultaneously subjected to two perpendicular simple harmonic motions represented by; x = a1 cos $\omega$t and y = a2 cos 2$\omega$t traces a curve given by:

The angular frequency of the damped oscillator is given by,

$\omega = \sqrt{\left(\frac{k}{m}-\frac{r^2}{4m^2}\right)}$ where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio $\frac{r^2}{mk}$ is 8%, the change in time period compared to the undamped oscillator is approximately as follows:

The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from 10 cm to 8 cm in 40 seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is 1.3, the time in which amplitude of this pendulum will reduce from 10 cm to 5 cm in carbon dioxide will be close to (ln 5 = 1.609, ln 2 = 0.693).

Bob of a simple pendulum of length $l$ is made of iron. The pendulum is oscillating over a horizontal coil carrying direct current. If the time period of the pendulum is T then:

The amplitude of a damped oscillator decreases to $0.9$ times its original magnitude in $5\text{ s}$. In another $10\text{ s}$ it will decrease to $\alpha$ times its original magnitude, where $\alpha$ equals

If a simple pendulum has significant amplitude (up to a factor of $1/e$ of original) only in the period between $t = 0\text{ s to } t = \tau\text{ s},$ then $\tau$ may be called the average life of the pendulum When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity with $b$ as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds:

In forced oscillation of a particle the amplitude is maximum for a frequency $\omega_1$ of the force while the energy is maximum for a frequency $\omega_2$ of the force; then

A particle of mass $m$ is attached to a spring (of spring constant $k$) and has a natural angular frequency $\omega_0.$ An external force $F(t)$ proportional to $\cos\omega t\ (\omega \neq \omega_0)$ is applied to the oscillator. The time displacement of the oscillator will be proportional to

In case of a forced vibration, the resonance peak becomes very sharp when the

When an oscillator completes 100 oscillations its amplitude is reduced to $\frac{1}{3}$ of initial value. What will be its amplitude, when it completes 200 oscillations?

A particle, with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force $F\sin\omega t$. If the amplitude of the particle is maximum for $\omega = \omega_1$ and the energy of the particle is maximum for $\omega = \omega_2$, then

(where $\omega_0 \rightarrow$ natural angular frequency of oscillations of particle)

Two SHM’s with same amplitude and time period, when acting together in perpendicular directions with a phase difference of $\pi/2$, give rise to

The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of $\pi$ results in the displacement of the particle along

A particle, with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force $F\sin\omega t$. If the amplitude of the particle is maximum for $\omega = \omega_1$ and the energy of the particle is maximum for $\omega = \omega_2$, then