Questions: Oscillations

Q1. Define simple harmonic motion and write its equation. Distinguish it from periodic motion.

Simple harmonic motion is a to and fro motion about a mean position in which the acceleration is proportional to the displacement from the mean position and directed towards it: a = -omega squared x. The displacement is a sinusoidal function of time, x = A cos(omega t + phi). Every SHM is periodic, but a periodic motion need not be simple harmonic: the motion of a clock hand or the rotation of the earth repeats at fixed intervals without a restoring force proportional to displacement. In SHM the period is independent of the amplitude, while in general periodic motion that need not hold.

Hinglish: SHM me acceleration displacement ke proportional aur opposite hoti hai: a = -ω²x. Displacement sinusoidal hota hai. Har SHM periodic hai, lekin har periodic motion SHM nahi hoti, jaise ghadi ki sui ya earth ka rotation.

Q2. Derive the expressions for velocity and acceleration in SHM and state where each is maximum.

Let x = A cos(omega t + phi). Differentiating once gives v = -A omega sin(omega t + phi). Using the displacement, this can be written as v = plus or minus omega sqrt(A squared minus x squared). Differentiating again gives a = -A omega squared cos(omega t + phi) = -omega squared x. The speed is maximum at the mean position, where x = 0 and v = A omega, and zero at the extremes. The acceleration vanishes at the centre and is largest in magnitude at the turning points, where it equals omega squared A. The velocity and the acceleration are always in opposite directions on the same side of the mean position.

Hinglish: x = A cos(ωt + φ) ko differentiate karo to v = -Aω sin(ωt + φ) milta hai. Mean position par speed max (Aω) aur extremes par zero hoti hai. Acceleration a = -ω²x hoti hai, jo mean par zero aur extremes par max hoti hai.

Q3. Show that the total energy of a simple harmonic oscillator is constant and proportional to the square of the amplitude.

The kinetic energy is (1/2) m v squared and the potential energy is (1/2) k x squared. With v squared = omega squared (A squared minus x squared) and k = m omega squared, the kinetic energy becomes (1/2) m omega squared (A squared minus x squared). Adding the potential energy gives E = (1/2) m omega squared A squared, which is independent of x and therefore constant. In terms of the force constant, E = (1/2) k A squared, so the energy depends on the square of the amplitude. Doubling the amplitude quadruples the energy.

Hinglish: K = (1/2)mv² aur U = (1/2)kx² ko jodo, to E = (1/2)kA² nikalta hai, jo x par depend nahi karta. Energy amplitude ke square ke proportional hoti hai, isliye amplitude double karne par energy chaar guna ho jaati hai.

Q4. A spring of force constant 200 N/m carries a 0.5 kg mass. Find the period, and the maximum speed if the amplitude is 4 cm.

The angular frequency is omega = sqrt(k/m) = sqrt(200/0.5) = sqrt(400) = 20 rad s^-1. The period is 2 pi/omega = 2 pi/20 = 0.314 s. The maximum speed is A omega = 0.04 x 20 = 0.8 m s^-1, reached at the mean position. The total energy is (1/2) k A squared = 0.5 x 200 x 0.0016 = 0.16 J. The maximum acceleration is A omega squared = 0.04 x 400 = 16 m s^-2, reached at the extreme positions.

Hinglish: ω = sqrt(k/m) = 20 rad/s, period 0.314 s. Maximum speed Aω = 0.8 m/s mean position par. Energy (1/2)kA² = 0.16 J aur maximum acceleration 16 m/s² extremum par milti hai.

Q5. Why is the period of a simple pendulum independent of the mass of the bob?

For a small displacement the restoring force is the component of the weight along the arc, which is m g sin theta, and it is proportional to the mass. The acceleration is the force divided by the mass, so the mass cancels and the acceleration is g sin theta, independent of m. The equation of motion therefore contains only g and l, and the period 2 pi sqrt(l/g) is the same for bobs of different masses. If the bob is hollow or made of a different metal, the period does not change as long as the string length is the same.

Hinglish: Restoring force mg sinθ mass ke proportional hota hai, aur acceleration force/mass se aata hai, isliye mass cancel ho jaata hai. Equation me sirf g aur l bachte hain, isliye period T = 2π√(l/g) mass par depend nahi karta.

Q6. Explain the phase constant and how it is determined from initial conditions.

The phase constant phi in x = A cos(omega t + phi) fixes the position and the direction of motion at t = 0. Given the initial displacement x0 and the initial velocity v0, we get x0 = A cos phi and v0 = -A omega sin phi. Squaring and adding eliminates phi and gives the amplitude A = sqrt(x0 squared + (v0/omega) squared). Dividing the two equations gives tan phi = -v0/(omega x0), and the quadrant of phi is chosen from the signs of x0 and v0. Two oscillators with the same amplitude and frequency can be out of step if their phase constants differ.

Hinglish: Phase constant φ se pata chalta hai ki t = 0 par particle kahan hai aur kis direction me chal raha hai. x0 = A cosφ aur v0 = -Aω sinφ se A = sqrt(x0² + (v0/ω)²) aur tanφ = -v0/(ωx0) milta hai. Quadrant signs se choose karo.

Q7. A particle in SHM has amplitude 10 cm and period 2 s. Find the time it takes to go from the mean position to 5 cm.

The angular frequency is omega = 2 pi/T = pi rad s^-1. Taking x = A sin(omega t) with the particle at the mean position at t = 0, we need 5 = 10 sin(pi t), so sin(pi t) = 0.5 and pi t = pi/6. Hence t = 1/6 s, about 0.167 s. The particle reaches the extreme at t = 0.5 s, so one quarter of the period is 0.5 s and the first 5 cm takes one third of that quarter. The answer would be different if the motion started at an extreme, because the cosine form would put the phase differently.

Hinglish: ω = π rad/s. Mean se 5 cm tak jaane ke liye x = A sin(ωt) me 5 = 10 sin(πt) rakho, to sin(πt) = 0.5 aur t = 1/6 s. Extreme tak jaane me 0.5 s lagta hai, isliye 5 cm pehle aata hai.

Q8. Compare the motion of a spring-mass system on the earth and on the moon.

For a spring-mass system the period is 2 pi sqrt(m/k). The spring constant depends on the spring itself, and the mass is the same on the earth and on the moon, so the period is unchanged. What changes is the equilibrium extension: on the moon the weight of the mass is about one sixth, so the spring stretches less, and the mass oscillates about a different mean position. For a pendulum, by contrast, the period contains g, so the same pendulum swings more slowly on the moon. This contrast is a favourite question: the spring period is independent of g, while the pendulum period is not.

Hinglish: Spring ka period 2π√(m/k) hota hai, isliye moon par bhi wahi rehta hai; sirf equilibrium stretch kam hota hai. Pendulum ka period 2π√(l/g) me g aata hai, isliye moon par pendulum dheere swing karta hai.

Q9. What is the physical significance of the force constant k?

The force constant measures the stiffness of the oscillator. A large k means a strong restoring force for a given displacement, so the motion is faster and the period is shorter. For a spring, k is the force needed to stretch it by one metre. The angular frequency is omega = sqrt(k/m), so the same spring oscillates more slowly with a heavier mass. In the energy expression the total energy is (1/2) k A squared, so a stiffer spring stores more energy for the same amplitude. The constant depends on the geometry and material of the oscillator, not on the motion.

Hinglish: Force constant k stiffness batata hai: bada k matlab zyada restoring force aur chhota period. ω = sqrt(k/m), isliye bhaari mass ke saath oscillation dheema hota hai. Energy (1/2)kA² bhi k se badhti hai. k oscillator ki property hai, motion par depend nahi karta.

Q10. Explain why the pendulum period increases for large amplitudes.

The derivation of the pendulum period uses sin theta approximately equals theta, which is accurate only for small angles. For larger angles, sin theta is smaller than theta, so the restoring force is weaker than the linear approximation predicts. A weaker restoring force means a smaller acceleration for the same displacement, so the bob takes longer to complete a swing and the period rises. The correction is small at 10 degrees and becomes noticeable at 30 degrees and beyond. This is why a grandfather clock is adjusted for a small amplitude, and why the simple formula is taught as a small-angle result.

Hinglish: Pendulum ka formula sinθ ≈ θ par based hai, jo sirf chhote angle ke liye theek hai. Bade angle par sinθ < θ, isliye restoring force kamzor ho jaata hai aur period badh jaata hai. Isliye ghadi ki pendulum chhoti amplitude par chalai jaati hai.

Quiz below: five questions on the SHM equation, velocity and acceleration, energy, the spring period and the pendulum.