Notes : Potential Energy of a Spring, Hooke's Law

Potential Energy of a Spring, Hooke's Law and Conservation of Mechanical Energy

1. Hooke's Law and Spring Force

An ideal spring exerts a restoring force that is directly proportional to its displacement from the equilibrium position and acts in the opposite direction.

Hooke's Law:

\[ F_s=-kx \]

Where:

• \(F_s\) = Spring force
• \(k\) = Spring constant
• \(x\) = Displacement from equilibrium position

Spring Constant (k)

• Large \(k\) → Stiff spring
• Small \(k\) → Soft spring

SI Unit:

\[ \text{N m}^{-1} \]

Meaning of the Negative Sign

The negative sign indicates that the spring force always acts towards the equilibrium position.

• If x>0 , if Spring is Stretched and Force acts Backward.
• If x < 0 , if Spring is Compressed, Force acts Forward.

Therefore, spring force is called a restoring force.

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2. Work Done by the Spring Force

Since spring force varies with displacement, it is a variable force. Therefore, work is calculated using integration.

\[ W=\int F\,dx \]

Substituting Hooke's Law:

\[ F=-kx \]

\[ W_s=\int_{x_i}^{x_f}(-kx)\,dx \]

Since \(k\) is constant, it can be taken outside the integral.

\[ W_s=-k\int_{x_i}^{x_f}x\,dx \]

Integration Without Skipping Steps

Using the power rule of integration:

\[ \int x^n\,dx=\frac{x^{n+1}}{n+1}+C \]

Here,

\[ n=1 \]

Therefore,

\[ \int x^1\,dx \]

\[ =\frac{x^{1+1}}{1+1} \]

\[ =\frac{x^2}{2} \]

Substituting into the work equation:

\[ W_s=-k\left[\frac{x^2}{2}\right]_{x_i}^{x_f} \]

Applying limits:

\[ W_s=-k\left(\frac{x_f^2}{2}-\frac{x_i^2}{2}\right) \]

Taking \(\frac{k}{2}\) common:

\[ W_s=-\frac{k}{2}(x_f^2-x_i^2) \]

Multiplying by negative sign:

\[ W_s=\frac{k}{2}(x_i^2-x_f^2) \]

Hence,

\[ \boxed{W_s=\frac12k(x_i^2-x_f^2)} \]

Special Case: Stretching from 0 to \(x_m\)

\[ x_i=0 \]

\[ x_f=x_m \]

Substituting:

\[ W_s=\frac12k(0^2-x_m^2) \]

Since

\[ 0^2=0 \]

Therefore,

\[ W_s=\frac12k(0-x_m^2) \]

\[ W_s=-\frac12kx_m^2 \]

\[ \boxed{W_s=-\frac12kx_m^2} \]

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3. Why is Spring Force a Conservative Force?

For motion from \(x_i\) to \(x_f\):

\[ W_1=\frac12k(x_i^2-x_f^2) \]

For return motion:

\[ W_2=\frac12k(x_f^2-x_i^2) \]

Total work:

\[ W_{total}=W_1+W_2 \]

\[ =\frac12k(x_i^2-x_f^2)+\frac12k(x_f^2-x_i^2) \]

Taking \(\frac12k\) common:

\[ W_{total} = \frac12k [x_i^2-x_f^2+x_f^2-x_i^2] \]

\[ =\frac12k(0) \]

\[ W_{total}=0 \]

Therefore, spring force is a conservative force.

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4. Potential Energy of a Spring

The relation between work and potential energy is:

\[ W_s=-(V_f-V_i) \]

Choose equilibrium position as reference:

\[ V(0)=0 \]

For displacement \(x\):

\[ W_s=-\frac12kx^2 \]

Substituting:

\[ -\frac12kx^2=-(V-0) \]

\[ -\frac12kx^2=-V \]

Multiplying both sides by \((-1)\):

\[ V=\frac12kx^2 \]

Hence,

\[ \boxed{V=\frac12kx^2} \]

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5. Verification Using Differentiation (No Steps Skipped)

For a conservative force:

\[ F=-\frac{dV}{dx} \]

Substitute:

\[ V=\frac12kx^2 \]

\[ F=-\frac{d}{dx}\left(\frac12kx^2\right) \]

Take constants outside:

\[ F=-\frac12k\frac{d}{dx}(x^2) \]

Using differentiation rule:

\[ \frac{d}{dx}(x^n)=nx^{n-1} \]

Here:

\[ n=2 \]

Therefore,

\[ \frac{d}{dx}(x^2) \]

\[ =2x^{2-1} \]

\[ =2x \]

Substituting:

\[ F=-\frac12k(2x) \]

\[ F=-\frac{2kx}{2} \]

\[ F=-kx \]

Hence,

\[ \boxed{F=-kx} \]

which verifies Hooke's Law.

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6. Conservation of Mechanical Energy

Total mechanical energy:

\[ E=K+V \]

\[ E=\frac12mv^2+\frac12kx^2 \]

Therefore,

\[ \boxed{\frac12mv^2+\frac12kx^2=\text{constant}} \]

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7. Energy at Important Positions

(A) Maximum Displacement (\(x=\pm x_m\))

At extreme position:

\[ v=0 \]

\[ K=\frac12m(0)^2 \]

\[ K=0 \]

\[ V=\frac12kx_m^2 \]

\[ E=\frac12kx_m^2 \]

Entire energy is potential energy.

(B) Equilibrium Position (\(x=0\))

\[ V=\frac12k(0)^2 \]

\[ V=0 \]

\[ E=K \]

\[ E=\frac12mv_m^2 \]

Entire energy is kinetic energy.

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8. Derivation of Maximum Speed

At maximum displacement:

\[ E=\frac12kx_m^2 \]

At equilibrium:

\[ E=\frac12mv_m^2 \]

By conservation of energy:

\[ \frac12kx_m^2=\frac12mv_m^2 \]

Multiplying both sides by 2:

\[ kx_m^2=mv_m^2 \]

Dividing by \(m\):

\[ v_m^2=\frac{kx_m^2}{m} \]

\[ v_m^2=\frac{k}{m}x_m^2 \]

Taking square root:

\[ v_m=\sqrt{\frac{k}{m}x_m^2} \]

\[ v_m=x_m\sqrt{\frac{k}{m}} \]

Hence,

\[ \boxed{v_m=x_m\sqrt{\frac{k}{m}}} \]

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9. Frequently Asked Questions (FAQ)

Q1. What is Hooke's Law?

Hooke's Law states that the restoring force of a spring is directly proportional to displacement and opposite in direction.

Q2. Why is there a negative sign in Hooke's Law?

The negative sign indicates that the force acts opposite to displacement.

Q3. What is the SI unit of spring constant?

\(\text{N m}^{-1}\)

Q4. Why is spring force conservative?

Because work done over a closed path is zero.

Q5. What is the potential energy of a spring?

\(\frac12kx^2\)

Q6. At which position is potential energy maximum?

At maximum extension and maximum compression.

Q7. At which position is kinetic energy maximum?

At equilibrium position.

Q8. What is the maximum speed of a spring-block system?

\(v_m=x_m\sqrt{\frac{k}{m}}\)

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10. MCQ Quiz

1. Hooke's Law is represented by:

   A) \(F=kx\)
   B) \(F=-kx\)
   C) \(F=mx\)
   D) \(F=\frac{k}{x}\)

   Answer: B

2. The SI unit of spring constant is:

   A) J
   B) N
   C) N·m
   D) N·m^-1

   Answer: D

3. Potential energy stored in a spring is:

   A) \(kx\)
   B) \(kx^2\)
   C) \(\frac12kx^2\)
   D) \(\frac{k}{x}\)

   Answer: C

4. At equilibrium position:

   A) PE is maximum
   B) KE is zero
   C) KE is maximum
   D) Total energy is zero

   Answer: C

5. Total mechanical energy of an ideal spring-mass system:

   A) Increases
   B) Decreases
   C) Remains constant
   D) Becomes zero

   Answer: C