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CBSE Class 11 Physics · 10 questions · 24 marks
Anything that is nudged away from a stable position and pushed back proportionally to how far it has strayed will oscillate in the same universal way. That single idea, simple harmonic motion, describes a swinging pendulum, a loaded spring, a vibrating tuning fork and the atoms in a crystal. In this chapter you will learn to write the equation of such motion, find its period, track the endless exchange between kinetic and potential energy, and see what happens when damping and driving forces are added.
A body executes simple harmonic motion of amplitude A. At what displacement from the mean position are its kinetic and potential energies equal?
Answer
A/√2 — the potential energy is ½mω²x² and the kinetic energy is ½mω²(A² − x²). Setting them equal gives x² = A² − x², so 2x² = A² and x = A/√2 ≈ 0.71A. At that point each accounts for half the total energy ½mω²A². Note that this happens at 71% of the amplitude, not at the midpoint, because the energies vary with the square of the displacement.
A simple pendulum has period T on the Earth's surface. Its period on a planet where the acceleration due to gravity is one-quarter of the Earth's value would be:
Answer
2T — since T = 2π√(L/g), the period varies as 1/√g. Reducing g to g/4 multiplies the period by √4 = 2. The mass of the bob is irrelevant, and the length is unchanged, so the pendulum swings exactly twice as slowly. This is also why a pendulum clock taken to a high mountain, where g is slightly smaller, runs slow.
The displacement of a particle is given by x = 0.05 sin(20t + π/3) metres, with t in seconds. Its maximum speed is:
Answer
1.0 m/s — comparing with x = A sin(ωt + φ) gives A = 0.05 m and ω = 20 rad/s. The velocity is v = Aω cos(ωt + φ), whose maximum magnitude is v_max = Aω = 0.05 × 20 = 1.0 m/s. This maximum occurs as the particle passes through the mean position, where the restoring force is momentarily zero.
Two identical springs, each of force constant k, are joined end to end and a mass m is hung from the combination. The period of vertical oscillation is:
Answer
2π√(2m/k) — springs joined end to end are in series, and both carry the same tension while their extensions add, so 1/k_eff = 1/k + 1/k = 2/k, giving k_eff = k/2. Then T = 2π√(m/k_eff) = 2π√(2m/k). A series combination is always softer than either spring alone, so the period lengthens; had the springs been side by side in parallel, k_eff would be 2k and the period would shorten.
Assertion (A): The time period of a simple pendulum does not change when the mass of the bob is doubled. Reason (R): The restoring force on the bob is proportional to its mass, so the mass cancels out when the acceleration is calculated.
Answer
Both A and R are true and R is the correct explanation of A — the tangential restoring force on the bob is mg sin θ, which is proportional to m. The resulting acceleration is a = F/m = g sin θ, from which the mass has vanished entirely. That is why T = 2π√(L/g) contains no mass term at all. The situation is exactly parallel to free fall, where all bodies accelerate equally for the same reason. Contrast this with a spring-mass system, where the restoring force −kx does not scale with m, so there the period does depend on the mass.
A block of mass 0.40 kg attached to a spring of force constant 160 N/m oscillates on a frictionless horizontal surface with an amplitude of 0.06 m. Find the period of oscillation and the total energy of the system.
Answer
Period: T = 2π√(m/k) = 2π√(0.40/160) = 2π√(2.5×10⁻³). √(2.5×10⁻³) = 0.05. T = 2π × 0.05 = 0.1π ≈ 0.314 s. Total energy: E = ½kA² = ½ × 160 × (0.06)² = 80 × 3.6×10⁻³ = 0.288 J. So the block completes about 3.2 oscillations each second, carrying a constant total energy of about 0.29 J that shuttles endlessly between kinetic and potential form.
A particle performs SHM with an amplitude of 0.10 m and a period of 0.40 s. Find (a) its maximum acceleration and (b) its speed when it is 0.06 m from the mean position. Take π² ≈ 9.87.
Answer
First find the angular frequency: ω = 2π/T = 2π/0.40 = 5π ≈ 15.71 rad/s. (a) Maximum acceleration occurs at the extreme positions, where x = A: a_max = ω²A = (5π)² × 0.10 = 25π² × 0.10 = 25 × 9.87 × 0.10 ≈ 24.7 m/s². (b) Speed at displacement x: v = ω√(A² − x²) = 15.71 × √((0.10)² − (0.06)²) = 15.71 × √(0.01 − 0.0036) = 15.71 × √(0.0064) = 15.71 × 0.08 ≈ 1.26 m/s. For comparison the maximum speed, at the centre, is ωA = 15.71 × 0.10 ≈ 1.57 m/s, so at 60% of the amplitude the particle still retains 80% of its top speed.
(a) Define simple harmonic motion and derive expressions for the velocity and acceleration of a particle executing SHM, given x = A sin(ωt + φ). (b) Derive the expression for the period of a simple pendulum, stating clearly the approximation used and the conditions under which the result holds. (c) A pendulum that beats seconds (period 2 s) is to be built. Find its length, taking g = 9.8 m/s² and π² = 9.87.
Answer
(a) Definition: simple harmonic motion is the motion of a particle whose acceleration is directly proportional to its displacement from a fixed mean position and is always directed towards that position. In symbols, a = −ω²x, or equivalently F = −kx with ω = √(k/m). The minus sign is the essential feature: it says the force always pulls the particle back. Velocity: the displacement is x = A sin(ωt + φ). Differentiating with respect to time: v = dx/dt = Aω cos(ωt + φ). This is maximum in magnitude, v_max = Aω, when cos(ωt + φ) = ±1, that is when sin(ωt + φ) = 0 and the particle is at the mean position. It is zero at the extremes. To express v in terms of x, use cos²θ + sin²θ = 1. Since sin(ωt + φ) = x/A, cos(ωt + φ) = ±√(1 − x²/A²). Therefore v = ±Aω√(1 − x²/A²) = ±ω√(A² − x²). Acceleration: differentiating the velocity, a = dv/dt = −Aω² sin(ωt + φ) = −ω²x. This confirms the defining condition of SHM and shows the acceleration is greatest, ω²A in magnitude, at the extreme positions and zero at the centre — exactly where the speed is greatest. (b) Period of a simple pendulum. Let a bob of mass m hang from an inextensible string of length L and be displaced so that the string makes a small angle θ with the vertical. Two forces act: the weight mg downward and the tension T along the string. Resolve the weight into a component mg cos θ along the string, which is balanced by the tension and provides the centripetal force, and a component mg sin θ perpendicular to the string, which is the restoring force: F = −mg sin θ, the minus sign showing it acts towards the mean position. For small angles measured in radians, sin θ ≈ θ. This is the key approximation, and it is good to better than 1% for θ up to about 14°. Hence F ≈ −mgθ. If x is the arc displacement of the bob, then θ = x/L, so F ≈ −(mg/L)x. This has the form F = −kx with k = mg/L, so the motion is simple harmonic. The acceleration is a = F/m = −(g/L)x, and comparing with a = −ω²x gives ω² = g/L, so ω = √(g/L). The period is T = 2π/ω = 2π√(L/g). Conditions for validity: the angular amplitude must be small so that sin θ ≈ θ; the string must be light and inextensible; the bob must be small enough to be treated as a point mass; and air resistance and friction at the support must be negligible. For large amplitudes the period grows slightly and the motion is periodic but no longer strictly simple harmonic. (c) Length of a seconds pendulum. T = 2π√(L/g), so squaring both sides: T² = 4π²L/g L = gT²/(4π²) L = (9.8 × 2²)/(4 × 9.87) L = 39.2/39.48 L ≈ 0.993 m. So a seconds pendulum is very close to one metre long — a coincidence that once tempted scientists to define the metre as the length of such a pendulum, an idea abandoned because g varies from place to place.
(a) Show that the total mechanical energy of a particle in SHM is constant and equal to ½mω²A², and describe how the kinetic and potential energies vary with displacement. (b) A 0.25 kg block on a frictionless surface is attached to a spring of force constant 100 N/m and pulled 0.08 m from equilibrium before release. Find the angular frequency, total energy, the speed at x = 0.04 m, and the displacement at which the kinetic energy is three times the potential energy.
Answer
(a) Energy in SHM. Let the displacement be x = A sin(ωt + φ), so the velocity is v = Aω cos(ωt + φ). Kinetic energy: KE = ½mv² = ½mA²ω² cos²(ωt + φ). Using cos²(ωt + φ) = 1 − sin²(ωt + φ) = 1 − x²/A²: KE = ½mω²(A² − x²). Potential energy: the restoring force is F = −kx = −mω²x, so the work done against it in moving the particle from 0 to x is stored as potential energy: PE = ∫ from 0 to x of mω²x dx = ½mω²x². Total energy: E = KE + PE = ½mω²(A² − x²) + ½mω²x² = ½mω²A². Every term in x has cancelled, so E is independent of the displacement and therefore constant throughout the motion. Since k = mω², it can also be written E = ½kA². Variation with displacement: at the mean position x = 0 the energy is entirely kinetic, KE = ½mω²A², and the speed is greatest. At the extremes x = ±A the energy is entirely potential, PE = ½mω²A², and the particle is momentarily at rest. In between, both vary as parabolas in x — the PE curve opening upward from the centre, the KE curve opening downward — and their sum is a horizontal line. Note that both energies pass through a full cycle twice in each cycle of the motion, so they oscillate at twice the frequency of the displacement itself. Note also that E ∝ A², so doubling the amplitude quadruples the energy. (b) Numerical work. Angular frequency: ω = √(k/m) = √(100/0.25) = √400 = 20 rad/s. Total energy: the block is released from rest at 0.08 m, so A = 0.08 m. E = ½kA² = ½ × 100 × (0.08)² = 50 × 6.4×10⁻³ = 0.32 J. Speed at x = 0.04 m: v = ω√(A² − x²) = 20 × √((0.08)² − (0.04)²) = 20 × √(6.4×10⁻³ − 1.6×10⁻³) = 20 × √(4.8×10⁻³) = 20 × 0.0693 ≈ 1.39 m/s. (As a check, the maximum speed is ωA = 20 × 0.08 = 1.6 m/s, and 1.39 m/s is sensibly less.) Displacement where KE = 3 PE: ½mω²(A² − x²) = 3 × ½mω²x² A² − x² = 3x² A² = 4x² x = A/2 = 0.08/2 = 0.04 m. So at exactly half the amplitude the kinetic energy is three times the potential energy — meaning the energy is split 75% kinetic and 25% potential, which fits the fact that PE ∝ x² and (1/2)² = 1/4.
Read the following and answer the questions that follow: A physics club builds a vertical spring oscillator to demonstrate damping. A 0.50 kg pan hangs from a light spring and, when set going in still air, the system oscillates with an amplitude that falls to half its initial value in 30 s. The spring stretches by 0.10 m when the pan is first attached. Take g = 10 m/s² and ln 2 ≈ 0.693. (a) Find the force constant of the spring and the period of small vertical oscillations. (b) The amplitude of a damped oscillator falls as A(t) = A₀e^(−bt/2m). Find the damping constant b. (c) By what factor has the total energy of the oscillator fallen after 30 s? (d) The club then shakes the support up and down at various frequencies. Describe what they will observe as the driving frequency is swept through the natural frequency, and explain how the amount of damping affects what they see.
Answer
(a) Force constant from the initial stretch. At equilibrium the spring force balances the weight: kx₀ = mg k = mg/x₀ = (0.50 × 10)/0.10 = 50 N/m. Period of vertical oscillation about this new equilibrium position: T = 2π√(m/k) = 2π√(0.50/50) = 2π√(0.01) = 2π × 0.1 ≈ 0.628 s. (Note that hanging the pan shifts the equilibrium point but does not change the period, because gravity is constant and simply relocates the mean position.) (b) Damping constant. The amplitude halves in t = 30 s, so A = A₀/2: A₀/2 = A₀e^(−b×30/(2×0.50)) 1/2 = e^(−30b) Taking natural logarithms of both sides: −ln 2 = −30b b = 0.693/30 ≈ 0.0231 kg/s. (c) Fall in total energy. The energy of an oscillator is proportional to the square of the amplitude, E ∝ A². If the amplitude has fallen to half, then E/E₀ = (A/A₀)² = (1/2)² = 1/4. The energy has therefore dropped to one-quarter of its initial value, a fall by a factor of 4, or a loss of 75%. The missing energy has gone into the surrounding air and into internal friction in the spring, appearing finally as heat. (d) Sweeping the driving frequency. At driving frequencies well below or well above the natural frequency f₀ = 1/T ≈ 1.6 Hz, the pan will respond with only a small amplitude. As the driving frequency is brought closer to f₀, the amplitude grows steadily, reaching a sharp maximum at resonance, where the driver feeds energy into the oscillator in step with its own motion so that every push arrives at the right moment. Past f₀ the amplitude falls away again. The phase also shifts: well below resonance the pan moves nearly in step with the driver, at resonance it lags by a quarter of a cycle, and well above it moves nearly in opposition. Effect of damping: damping decides how tall and how narrow the resonance peak is. With light damping the peak is very high and very sharp, so the club must tune the driving frequency delicately to find it, and the resonant amplitude may become large enough to be dangerous. Adding damping — for instance by immersing the pan in a dish of oil — lowers the peak and broadens it, so a wider band of frequencies produces a moderate response and no single frequency produces a violent one. Damping also shifts the peak very slightly below f₀. This is exactly why engineers deliberately add dampers to bridges and tall buildings: not to prevent resonance, which cannot be avoided, but to keep its consequences finite.
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