Simple Harmonic Motion

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

If a simple pendulum has significant amplitude (up to a factor of $$1/e$$ of original ) only in the period between $$t = 0s\,\,to\,\,t = \tau \,s,$$ 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 :

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

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 :

A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be (Assume that the highest frequency a person can hear is 20,000 Hz)

A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10^–2 m. The relative change in the angular frequency of the pendulum is best given by :

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 :-

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 :

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 drop to half of its initial value? (ln 2 = 0.693)

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)

If a spring has time period $$T,$$ and is cut into $$n$$ equal parts, then the time period of each part will be

In a simple harmonic oscillator, at the mean position

The displacement of particle varies according to the relation $$x=4$$$$\left( {\cos \,\pi t + \sin \,\pi t} \right).$$ The amplitude of the particle is

A body executes simple harmonic motion. The potential energy $$(P.E),$$ the kinetic energy $$(K.E)$$ and total energy $$(T.E)$$ are measured as a function of displacement $$x.$$ Which of the following statements is true ?

The total energy of particle, executing simple harmonic motion is

The function $${\sin ^2}\left( {\omega t} \right)$$ represents

Two simple harmonic motions are represented by the equations $${y_1} = 0.1\,\sin \left( {100\pi t + {\pi \over 3}} \right)$$ and $${y_2} = 0.1\,\cos \,\pi t.$$ The phase difference of the velocity of particle $$1$$ with respect to the velocity of particle $$2$$ is

If a simple harmonic motion is represented by $${{{d^2}x} \over {d{t^2}}} + \alpha x = 0.$$ its time period is

A coin is placed on a horizontal platform which undergoes vertical simple harmonic motoin of angular frequency $$\omega .$$ The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time