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Oscillations - Force Law for Simple Harmonic Motion

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Simple Harmonic Motion (SHM) is defined as a type of periodic motion where the restoring force is directly proportional to the displacement of the particle from its mean position.

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The restoring force FF always acts in a direction opposite to the displacement xx, mathematically expressed as F=−kxF = -kx.

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The proportionality constant kk is known as the force constant or spring constant, representing the 'stiffness' of the restoring mechanism. Its SI unit is N⋅m−1N \cdot m^{-1}.

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According to Newton's Second Law, F=maF = ma. Therefore, for a particle of mass mm executing SHM, the acceleration aa is given by a=−kmxa = -\frac{k}{m}x.

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Comparing this with the standard SHM acceleration equation a=−ω2xa = -\omega^2 x, we find that the angular frequency ω\omega is related to the force constant by ω=km\omega = \sqrt{\frac{k}{m}}.

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The negative sign in the force law F=−kxF = -kx indicates that the force is a 'restoring force', always directed towards the equilibrium (mean) position (x=0x = 0).

📐Formulae

F=−kxF = -kx

a=−ω2xa = -\omega^2 x

ω=km\omega = \sqrt{\frac{k}{m}}

T=2πmkT = 2\pi \sqrt{\frac{m}{k}}

f=12πkmf = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

k=mω2k = m\omega^2

💡Examples

Problem 1:

A body of mass m=0.5 kgm = 0.5\text{ kg} is executing SHM with a force constant k=50 N/mk = 50\text{ N/m}. Calculate the angular frequency ω\omega and the time period TT of the oscillation.

Solution:

Given: m=0.5 kgm = 0.5\text{ kg}, k=50 N/mk = 50\text{ N/m}. Using the formula for angular frequency: ω=km=500.5=100=10 rad/s\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{0.5}} = \sqrt{100} = 10\text{ rad/s} Now, calculating the time period: T=2πω=2×3.1410=0.628 sT = \frac{2\pi}{\omega} = \frac{2 \times 3.14}{10} = 0.628\text{ s}

Explanation:

The angular frequency is determined by the ratio of the restoring force per unit displacement to the mass. The time period is the inverse of the frequency multiplied by 2π2\pi.

Problem 2:

A particle of mass 0.2 kg0.2\text{ kg} executes SHM. When the displacement is x=0.1 mx = 0.1\text{ m}, the restoring force is F=−2 NF = -2\text{ N}. Find the acceleration of the particle at this point.

Solution:

Given: m=0.2 kgm = 0.2\text{ kg}, x=0.1 mx = 0.1\text{ m}, F=−2 NF = -2\text{ N}. Using Newton's Second Law: a=Fma = \frac{F}{m} a=−20.2=−10 m/s2a = \frac{-2}{0.2} = -10\text{ m/s}^2 Alternatively, using the force law F=−kxF = -kx: k=−Fx=−−20.1=20 N/mk = -\frac{F}{x} = -\frac{-2}{0.1} = 20\text{ N/m} a=−kmx=−200.2×0.1=−10 m/s2a = -\frac{k}{m}x = -\frac{20}{0.2} \times 0.1 = -10\text{ m/s}^2

Explanation:

Acceleration in SHM is directly proportional to displacement. The calculation shows that at a displacement of 0.1 m0.1\text{ m}, the particle experiences an acceleration of 10 m/s210\text{ m/s}^2 directed towards the mean position.