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COMEDK Oscillations — practice questions

Explore 65 free Oscillations questions in COMEDK Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Oscillations questions and exercises with answers

Showing questions 1–30 of 65.

The time period of a simple pendulum of length $L$ in a gravitational field $g$ is:Read the answer and worked solution →A particle in SHM has amplitude $A$ and angular frequency $\omega$. Its maximum speed is:Read the answer and worked solution →The time period of a spring-mass system with mass $m$ and spring constant $k$ is:Read the answer and worked solution →The total mechanical energy of a particle of mass $m$ in SHM with amplitude $A$ and angular frequency $\omega$ is:Read the answer and worked solution →A particle moving back and forth over the same path is undergoing:Read the answer and worked solution →In SHM, the time for one complete oscillation is called:Read the answer and worked solution →The general equation for SHM is:Read the answer and worked solution →In SHM, displacement and acceleration are:Read the answer and worked solution →The angular frequency ω of a spring-mass system with mass m and spring constant k is:Read the answer and worked solution →Time period of a simple pendulum of length L (small oscillations):Read the answer and worked solution →A spring-mass system has mass 4 kg and spring constant 100 N/m. Frequency of SHM:Read the answer and worked solution →For SHM x = 5 sin(2πt), find amplitude and time period:Read the answer and worked solution →Maximum velocity of a particle in SHM with amplitude A and angular frequency ω:Read the answer and worked solution →Maximum acceleration of a particle in SHM (amplitude A, angular frequency ω):Read the answer and worked solution →A particle executing SHM has time period T. Time taken to go from extreme position (x = A) to x = A/2:Read the answer and worked solution →A simple pendulum of length 1 m has period T (g = 9.8). New length needed to double T:Read the answer and worked solution →Total energy of SHM particle equals:Read the answer and worked solution →At the extreme position in SHM, kinetic energy equals:Read the answer and worked solution →Two springs k1 and k2 in parallel: equivalent spring constant:Read the answer and worked solution →Two springs k1 and k2 in series: equivalent spring constant:Read the answer and worked solution →A spring of force constant k is cut into two equal halves. Spring constant of each half:Read the answer and worked solution →A particle executing SHM has time period T. Time taken to go from extreme to mid-amplitude position:Read the answer and worked solution →In SHM, when does kinetic energy equal potential energy?Read the answer and worked solution →A simple pendulum swings with period T on Earth. On Moon (g_moon = g/6), its period:Read the answer and worked solution →For SHM x = A sin(ωt), find ratio of kinetic energy at t = T/4 to t = T/8:Read the answer and worked solution →A particle's SHM is described by x = 3 sin(ωt) + 4 cos(ωt). Amplitude of resulting SHM:Read the answer and worked solution →Time period of a simple pendulum is T. If mass is doubled, period becomes:Read the answer and worked solution →A particle of mass 0.5 kg has SHM with amplitude 4 cm and time period 0.2 s. Find maximum kinetic energy:Read the answer and worked solution →A simple pendulum of length 1 m on Earth (g = 10) has frequency:Read the answer and worked solution →For SHM, the displacement-time graph is:Read the answer and worked solution →

Worked practice questions

The time period of a simple pendulum of length $L$ in a gravitational field $g$ is:

  1. $T = 2\pi \sqrt{\dfrac{L}{g}}$
  2. $T = 2\pi \sqrt{\dfrac{g}{L}}$
  3. $T = 2\pi L g$
  4. $T = \sqrt{\dfrac{L}{g}}$
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Correct answer: A. $T = 2\pi \sqrt{\dfrac{L}{g}}$

For small oscillations the pendulum's equation of motion reduces to SHM with $\omega = \sqrt{g/L}$. So period $T = 2\pi/\omega = 2\pi\sqrt{L/g}$. Independent of the mass of the bob (a key contrast with the spring-mass system, where $T \propto \sqrt{m}$). Note: only valid for small-amplitude swings. At larger amplitudes the small-angle approximation fails and the period depends slightly on amplitude.

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A particle in SHM has amplitude $A$ and angular frequency $\omega$. Its maximum speed is:

  1. $A\omega$
  2. $A\omega^2$
  3. $A/\omega$
  4. $\omega/A$
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Correct answer: A. $A\omega$

Position $x(t) = A \sin(\omega t + \phi)$. Velocity is its time derivative: $v(t) = A\omega \cos(\omega t + \phi)$. Maximum value is $v_{\max} = A\omega$, achieved when the particle passes through the mean position ($x = 0$). Maximum acceleration is at the extremes: $a_{\max} = A\omega^2$ (option B is the acceleration max, the common trap).

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The time period of a spring-mass system with mass $m$ and spring constant $k$ is:

  1. $T = 2\pi \sqrt{\dfrac{k}{m}}$
  2. $T = 2\pi \sqrt{\dfrac{m}{k}}$
  3. $T = 2\pi k m$
  4. $T = 2\pi (k + m)$
Show answer and solution

Correct answer: B. $T = 2\pi \sqrt{\dfrac{m}{k}}$

Spring-mass system: Newton's second law gives $m\ddot x = -kx$, hence $\omega = \sqrt{k/m}$. Period: $T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{m/k}$. A stiffer spring (larger $k$) gives a shorter period. A heavier mass gives a longer period. Option A inverts the ratio, the classic trap.

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