#harmonic oscillator

Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.8 \, \text{kg}, k = 320 \, \text{N/m} \). If displaced by \( 5 \, \text{cm} \), what i

**Mass-spring dynamics** show T depends on mass and stiffness, independent of amplitude for ideal spring. Given T and m, k = 4π² m/T² extracted, and energy E = ½ k A² connects amplitude to total mechanical energy, illustrating isochronism. Total energy: E = (1/2) k A² . A = 0.05 m, k = 320 N/m . E = 0.5 × 320 × (0.05)² = 0.5 × 320 × 0.0025 = 0.4 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.4 J follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Spring-Mass System and Combination of Springs

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Equations of SHM, Phase and Angular Frequency

A particle in SHM has \( a = -9 x \) (in SI units). What is its period?

**Angular frequency** ω = 2πf = 2π/T characterizes rapidity, independent of amplitude for SHM. Phase constant φ shifts sine/cosine, allowing any initial condition, e.g., x(0)=0 requires φ=0 for sine form. Displacement, velocity, acceleration share ω but differ in phase by 90° and 180°. For SHM, a = -ω² x . Given a = -9 x , ω² = 9 ⇒ ω = 3 rad/s . T = (2π/ω) = (2π/3) ≈ 2.09 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.09 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

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Periodic Motion and SHM Basic Concepts

A particle’s acceleration in SHM is \( a = -25 x \) (in SI units). What is its frequency?

**Periodic motion** repeats after fixed period T, x(t+T)=x(t), while oscillatory motion involves to-and-fro about equilibrium. SHM is special periodic motion where restoring force proportional to displacement, F = -k x, acceleration a = -ω² x, leading to sinusoidal displacement x = A cos(ωt + φ). For SHM, a = -ω² x . Given a = -25 x , ω² = 25 ⇒ ω = 5 rad/s . Frequency: v = (ω/2π) = (5/2 × 3.14) ≈ 0.796 Hz . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.796 Hz follows,

Ref: NCERT > Physics Book > Oscillations > Periodic Motion and SHM Basic Concepts

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Periodic Motion and SHM Basic Concepts

A particle in SHM has \( a = -100 x \) (in SI units). What is its frequency?

**SHM condition** is linear restoring force and inertia producing sinusoidal time dependence. Motions violating a = -ω² x, such as uniform circular motion, are periodic without oscillation about fixed point, highlighting classification criteria for NCERT. For SHM, a = -ω² x . Given a = -100 x , ω² = 100 ⇒ ω = 10 rad/s . Frequency: v = (ω/2π) = (10/2 × 3.14) ≈ 1.59 Hz . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.59 Hz follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Periodic Motion and SHM Basic Concepts

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Energy in SHM - Kinetic, Potential and Total

A spring system has \( m = 1.2 \, \text{kg}, k = 480 \, \text{N/m}, A = 6 \, \text{cm} \). What is the potential energy

**Energy in SHM** interconverts between kinetic K = ½ m v² = ½ m ω² (A² - x²) and potential U = ½ k x² = ½ m ω² x², total E = K + U = ½ k A² = ½ m ω² A² constant, independent of time. At mean position x=0, E = K_max = ½ m ω² A², at extremes x=±A, E = U_max = ½ k A². Potential energy: U = (1/2) k x² . k = 480 N/m, x = 0.03 m . U = 0.5 × 480 × (0.03)² = 0.5 × 480 × 0.0009 = 0.216 J .

Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total

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