AP EAMCET Oscillations — practice questions
60 free MCQs with worked solutions. Tap any question for the answer + explanation, or practice them all in the app.
Practice AP EAMCET Oscillations in the app →A particle executes simple harmonic motion (SHM) about its mean position. The restoring force on it is:The time period of a simple pendulum of length $L$ in a gravitational field $g$ is:A particle in SHM has amplitude $A$ and angular frequency $\omega$. Its maximum speed is:The time period of a spring-mass system with mass $m$ and spring constant $k$ is:The total mechanical energy of a particle of mass $m$ in SHM with amplitude $A$ and angular frequency $\omega$In a damped oscillation, the amplitude decreases exponentially as $A(t) = A_0 e^{-\gamma t}$. The time after wThe time period of a simple pendulum of length $L$ is:Time period of a spring-mass system (spring constant $k$, mass $m$) is:At the **mean (equilibrium) position** of SHM, which is true?Total mechanical energy of a particle in SHM (mass $m$, amplitude $A$, angular frequency $\omega$) is:If displacement of a particle in SHM is $x = A in(\omega t)$, velocity at $x = A/2$ is:In SHM, when the displacement is half the amplitude (x = A/2), the ratio of KE to PE is:A simple pendulum of period $T$ on Earth is taken to the Moon where $g_{moon} = g/6$. Its new period is:The phase difference between displacement $x$ and velocity $v$ in SHM is:A spring of force constant $k$ is cut into two equal halves. The force constant of each half is:Two springs of constants $k_1$ and $k_2$ connected in **series** have effective spring constant:If $x_1 = A in(\omega t)$ and $x_2 = A in(\omega t + \pi/3)$, the resultant amplitude of $x_1 + x_2$ is:Maximum acceleration in SHM ($A = $ amplitude, $\omega = $ angular frequency) is:In a damped oscillator $\ddot x + 2\gamma\dot x + \omega_0^2 x = 0$, the **amplitude decays** as:A particle in SHM has displacement $x = a\cos(\omega t) + b in(\omega t)$. The amplitude of motion is:A simple pendulum oscillates in a lift accelerating UPWARD with acceleration $a$. The effective time period beTwo identical pendulums of length $L$ are connected by a light spring. They oscillate **in opposite phases** (A block of mass $m$ on a smooth horizontal surface is attached to two walls by springs of constant $k_1$ and $A particle performs SHM along x-axis with amplitude $A$ and period $T$. The time taken from $x = 0$ to $x = A/A particle in SHM has a maximum speed of 30 cm/s and a maximum acceleration of 90 cm/s². The period of oscillaA spring of natural length $L$ has spring constant $k$. If it is cut into two pieces of lengths in ratio $1:2$A liquid is filled to height $h$ in a U-tube of uniform cross-section. The liquid column is displaced and releAn SHM is given by $x = 5 in(4\pi t + \pi/4)$ cm. The displacement at $t = 0.25$ s is:A mass $m$ on a vertical spring stretches it by $x_0$ when in equilibrium. The time period of small oscillatioA particle moving back and forth over the same path is undergoing:In SHM, the time for one complete oscillation is called:The general equation for SHM is:In SHM, displacement and acceleration are:The angular frequency ω of a spring-mass system with mass m and spring constant k is:Time period of a simple pendulum of length L (small oscillations):A spring-mass system has mass 4 kg and spring constant 100 N/m. Frequency of SHM:For SHM x = 5 sin(2πt), find amplitude and time period:Maximum velocity of a particle in SHM with amplitude A and angular frequency ω:Maximum acceleration of a particle in SHM (amplitude A, angular frequency ω):A particle executing SHM has time period T. Time taken to go from extreme position (x = A) to x = A/2:A simple pendulum of length 1 m has period T (g = 9.8). New length needed to double T:Total energy of SHM particle equals:At the extreme position in SHM, kinetic energy equals:Two springs k1 and k2 in parallel: equivalent spring constant:Two springs k1 and k2 in series: equivalent spring constant:A spring of force constant k is cut into two equal halves. Spring constant of each half:A particle executing SHM has time period T. Time taken to go from extreme to mid-amplitude position:In SHM, when does kinetic energy equal potential energy?A simple pendulum swings with period T on Earth. On Moon (g_moon = g/6), its period:For SHM x = A sin(ωt), find ratio of kinetic energy at t = T/4 to t = T/8:A particle's SHM is described by x = 3 sin(ωt) + 4 cos(ωt). Amplitude of resulting SHM:Time period of a simple pendulum is T. If mass is doubled, period becomes:A particle of mass 0.5 kg has SHM with amplitude 4 cm and time period 0.2 s. Find maximum kinetic energy:A simple pendulum of length 1 m on Earth (g = 10) has frequency:For SHM, the displacement-time graph is:A mass m on a vertical spring of constant k. The equilibrium extension is x0 = mg/k. The mass is set into SHM.Two simple pendulums of lengths 1 m and 4 m start swinging in phase. They will be in phase again after:A particle has SHM with x = A cos(ωt). At t = 0, the particle is at:Phase difference between displacement and velocity in SHM:The total energy of SHM is proportional to: