Oscillations

Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum L, notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as : (Take π² = 9.8, and g = 9.8 m/s²)

The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum bob at equilibrium position is approximately : (Consider mass of the bob = 20 g)

The two-dimensional motion of a particle, described by \(\vec{\mathrm{r}}=(\hat{i}+2\hat{j})\mathrm{Acos}\omega t\) 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 :

If the mass of the bob in a simple pendulum is increased to thrice to its original mass and its length is made half its original length, then the new time period of oscillations \(\frac{x}{2}\) times its original time period. Find the value of x:

The two-dimensional motion of a particle, described by \(\vec{\mathrm{r}}=(\hat{i}+2\hat{j})\mathrm{Acos}\omega t\) 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 :

[Re-NEET 2024]

A particle executing simple harmonic motion with amplitude A has the same potential and kinetic energies at the displacement :

[Re-NEET 2024]

If \(x=5\sin \left(\pi t+\frac{\pi }{3}\right)m\) represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are :

[NEET 2024]

Identify the function which represents a periodic motion.

[Re-NEET 2020]

A particle starts executing simple harmonic motion (SHM) of amplitude ' \(a\) ' and total energy \(E\). At any instant, its kinetic energy is \(3 E / 4\) then its displacement ' \(y\) ' is given by:

\(T_0\) is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to \(\frac{1}{16}\) times of its initial value, the modified time period is:

[JEE Main 2021, 22 Jul (Shift 2)]

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet.
Reason (R): The mass of the pendulum remains unchanged at Earth and the other planet.

In the light of the above statements, choose the correct answer from the options given below :

Given below are two statements :

Statement-I: A second's pendulum has a time period of 1 second.
Statement-II: It takes precisely one second to move between the two extreme positions in a second's pendulum.
In the light of the above statements, choose the correct answer from the options given below:

\(Y=A \sin \left(\omega t +\phi_0\right)\) is the time-displacement equation of a SHM. At \(t=0\) the displacement of the particle is \(Y=\frac{A}{2}\) and it is moving along negative \(x\)-direction. Then the initial phase angle \(\phi_0\) will be:

\(Y=A \sin \left(\omega t +\phi_0\right)\) is the time-displacement equation of a SHM. At \(t=0\) the displacement of the particle is \(Y=\frac{A}{2}\) and it is moving along negative \(x\)-direction. Then the initial phase angle \(\phi_0\) will be:

[JEE Main 2021, 25 Feb (Shift 2)]

A particle is executing \( SHM \). Then, the graph of velocity as a function of displacement is

[JEE Main 2021, 26 Feb (Shift 2)]

Average velocity of a particle executing SHM in one complete vibration is

[NEET 2019]

A particle is executing Simple Harmonic Motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be:

[JEE Main 2023, 12 Apr (Shift 1)]

The phase difference between displacement and acceleration of a particle in a simple harmonic motion is

An object of mass \(0.5 kg\) executing simple harmonic motion. Its amplitude is \(5 cm\) and time period \((T)\) is \(0.2 s\). What will be the potential energy of the object at an instant \(t=\frac{T}{4} s\) starting from mean position. Assume that the initial phase of the oscillation is zero.

[JEE Main 2021, 27 Jul (Shift 2)]

A spring is stretched by \( 5 \mathrm{~cm} \) by a force \( 10 \mathrm{~N} \). The time period of the oscillations when a mass of \( 2 \mathrm{~kg} \) is suspended by it is:

[NEET 2021]

If two similar spring each of spring constant \(K_1\) are joined in series, the new spring constant and time period would be changed by a factor :

[JEE Main 2021, 26 Feb (Shift 1)]

Two particles \(A\) and \(B\) of equal masses are suspended from two massless springs constants \(k_{1}\) and \(k_{2}\), respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of \(A\) and \(B\) is

[JEE Main 2021, 17 Mar (Shift 2)]

The phase difference between displacement and acceleration of a particle in a simple harmonic motion is :

[NEET 2020]

\(T_0\) is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to \(\frac{1}{16}\) times of its initial value, the modified time period is:

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 \(\ln 2=0.693\) )

[JEE Main 2021, 17 Mar (Shift 2)]

A body is executing simple harmonic motion with frequency \( 'n' \) , the frequency of its potential energy is:

[NEET 2021]

Two simple harmonic motions are represented by the equations

\({x}_{1}=5\sin \left(2\mathrm{πt}+\frac{\pi }{4}\right)\mathrm{and}{x}_{2}=5\sqrt{2}\left(\sin 2\mathrm{πt}+\cos 2\mathrm{πt}\right)\)

The ratio of the amplitude of \(x_{1}\) and \(x_{2}\) is