iCON Education - Simple Harmonic Motion TS-EAMCET Test

The force ( $F$ in newton) acting on a particle of mass 90 g executing simple harmonic motion is given by $F+0.04 \pi^2 y=0$, where $y$ is displacement of the particle in metre. If the amplitude of the particle is $\frac{6}{\pi} \mathrm{~m}$, then the maximum velocity of the particle is

$6 \mathrm{~ms}^{-1}$

$2 \mathrm{~ms}^{-1}$

$8 \mathrm{~ms}^{-1}$

$4 \mathrm{~ms}^{-1}$

If the amplitudes of a damped harmonic oscillator at times $t=0, t_1$ and $t_2$ are $A_0, A_1$ and $A_2$ respectively, then the amplitude of the oscillator at a time of $\left(t_1+t_2\right)$ is

$\frac{A_0+A_1+A_2}{3}$

$\frac{A_2 A_0}{A_1}$

$\frac{A_1 A_0}{A_2}$

$\frac{A_1 A_2}{A_0}$

At a given place, to increase the number of oscillations made by a simple pendulum in one minute from 72 to 90 , the length of the pendulum is to be decreased by

If the amplitude of a damped harmonic oscillator becomes half of its initial amplitude in a time of 10 s , then the time taken for the mechanical energy of the oscillator to become half of its initial mechanical energy is

A particle is executing simple harmonic motion. If the force acting on the particle at a position is $86.6 \%$ of the maximum force on it, then the ratio of its velocity at that point and its maximum velocity is

The amplitude of a particle executing simple harmonic motion is 6 cm . The distance of the point from the mean position at which the ratio of the potential and kinetic energies of the particle becomes $4: 5$ is

In a simple pendulum experiment for the determination of acceleration due to gravity, the error in the measurement of the length of the pendulum is $1 \%$ and the error in the measurement of the time period is $2 \%$. The error in the estimation of acceleration due to gravity is

A massless spring of length $l$ and spring constant $k$ oscillates with a time period $T$ when loaded with a mass $m$. The spring is now cut into three equal parts and are connected in parallel. The frequency of oscillation of the combination when it is loaded with ${ }_{3}$ mass 4 m is

If a body dropped freely from a height of 20 m reaches the surface of a planet with a velocity of $31.4 \mathrm{~ms}^{-1}$. then the length of a simple pendulum that ticks seconds on the planet is

A particle of mass 4 mg is executing simple harmonic motion along $X$-axis with an angular frequency of $40 \mathrm{rad} \mathrm{s}^{-1}$. If the potential energy of the particle is $V(x)=a+b x^2$, where $V(x)$ is in joule and $x$ is in metre, then the value of $b$ is

$800 \times 10^{-6} \mathrm{Jm}^{-2}$

$1600 \times 10^{-6} \mathrm{Jm}^{-2}$

$3200 \times 10^{-6} \mathrm{Jm}^{-2}$

$6400 \times 1^{-6} \mathrm{Jm}^{-2}$

In a time $t$ amplitude of vibrations of a damped oscillator becomes half of its initial value, then the mechanical energy of the oscillator decreases by

The displacement of a particle is given by the relation $x=4(\cos \pi t+\sin \pi t)$. The amplitude of the particle is

The displacement of a particle executing simple harmonic motion is given by $x=2 \cos (t)$ where $t$ is the time in seconds then the time period of the particle is

A force of 6.4 N stretches a vertical spring by 0.1 m . If it were to oscillate with a period of $\pi / 4$, then the mass that is to be suspended from the spring is

$\frac{\pi}{4} \mathrm{~kg}$

$\frac{1}{\pi} \mathrm{~kg}$

A pendulum has a time period $T$ in air. Whạt it is made to oscillate in water its time period is $\sqrt{2} T$. Then the relative density of the material of the bob of the pendulum is (neglect damping)

A clock is designed based on the oscillation of a spring-block system suspended vertically in the absence of air-resistance. Assume it shows the correct time when a spring of stiffness $k$ and block is mass $m$ are used. If the block is replaced by another block of mass $4 m$, choose the correct option

The clock runs slow by 0.5 s for every second

The clock runs fast by 0.5 s for every one second

The clock runs fast by 1 s for every one second

The clock runs slow by 1 s for every one second

For a particle executing simple harmonic motion, the kinetic energy of the particle at a distance of 4 cm from the mean position is $1 / 3$ rd of the maximum kinetic energy. The amplitude of the motion is

$2 \sqrt{6} \mathrm{~cm}$

$\frac{2}{\sqrt{6}} \mathrm{~cm}$

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

$\frac{6}{\sqrt{2}} \mathrm{~cm}$

A block is in simple harmonic motion (SHM) on the end of the spring with position given by $x=5 \cos \left(\omega t+\frac{\pi}{4}\right) \mathrm{cm}$. If the total mechanical energy used is 100 J to achieve maximum displacement, then the potential energy at time, $t=0$ is

A particle performs simple harmonic motion with a time period of 16 s . At a time $t=2 \mathrm{~s}$, the particle passes through the origin and at $t=4 \mathrm{~s}$ its velocity is $4 \mathrm{~m} / \mathrm{s}$. The amplitude of the motion is

$\frac{32 \pi}{\sqrt{2}}$

$\frac{32 \sqrt{2}}{\pi}$

The amplitude of a damped oscillator varies with time as $A(t)=A_0 \exp (-b t / 2 \mathrm{~m})$, where $b=70 \mathrm{~g} / \mathrm{s}$ and $m=200$ g. How long does it take for the mechanical energy to drop to one-fourth of its initial value?

[Take, $\ln 2=0.7$ ]

A simple pendulum of length 1 m and having a bob of mass 100 g is suspended in a car, moving on a circular track of radius 100 m with uniform speed $10 \mathrm{~m} / \mathrm{s}$. If the pendulum makes small oscillation in a radial direction

about its equilibrium position, then its time period can be given by $T=2 \pi / \alpha^{1 / 4}$. The value of $\alpha$ is

[Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]

A simple pendulum consists of a small sphere of mass $m$ suspended by a thread of length $l$. The sphere carries a positive charge $q$. The pendulum is allowed to do small oscillations in uniform electric field $E$ with direction vertically upwards. The time period of oscillation is

$2 \pi \sqrt{\frac{1}{g}}$

$2 \pi \sqrt{\frac{m l}{q E}}$

$2 \pi \sqrt{\frac{1}{g+\frac{q}{m} E}}$

$2 \pi \sqrt{\frac{1}{g-\frac{q}{m} E}}$

A body starting at $t=0$ from origin oscillates simple harmonically with a period of 4 s . After what time will its kinetic energy by $75 \%$ of its total energy?

A stiff spring having spring constant $k=400 \mathrm{~N} / \mathrm{m}$ is attached to the floor vertically. A mass $m=10 \mathrm{~kg}$ is placed on top of the spring. The block oscillates if it is pressed downward and released. Find the extension in the spring at which the block loses contact with spring. (Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

TS EAMCET 2020 (Online) 14th September Evening Shift Physics - Simple Harmonic Motion Question 3 English

A particle is executing simple harmonic motion in one-dimension. If the amplitude of oscillations is 0.2 cm and if its velocity at the mean position is $5 \mathrm{~m} / \mathrm{s}$, then the angular frequency of the oscillation is

$1000 \mathrm{rad} / \mathrm{s}$

$1500 \mathrm{rad} / \mathrm{s}$

$2000 \mathrm{rad} / \mathrm{s}$

$2500 \mathrm{rad} / \mathrm{s}$

A body is oscillating in simple harmonic motion according to the equation $x=6 \cos \left(2 \pi t+\frac{\pi}{3}\right) \mathrm{m}$. The magnitude of the acceleration (in $\mathrm{m} / \mathrm{s}^2$ ) of the body at $t=\mathrm{ls}$

A point mass oscillates along the $X$-axis according to the law $x=x_0 \cos \left(\omega t-\frac{\pi}{4}\right)$. If the acceleration of the particle is written as $a=A \cos (\omega t-\delta)$, then

$A=x_0 \omega^2, \delta=\frac{-3 \pi}{4}$

$A=x_0, \delta=-\frac{\pi}{4}$

$A=x_0 \omega^2, \delta=\frac{\pi}{4}$

$A=x_0 \omega^2, \delta=\frac{3 \pi}{4}$

For a particle executing SHM, determine the ratio of average acceleration of the particle between extreme position and equilibrium position w.r.t. the maximum acceleration.