An object of mass 0.2 kg executes simple harmonic oscillations along the X-axis with a frequency of

Q: Why can we assume that the total energy is constant throughout the motion?

In simple harmonic motion, unless specified otherwise, we assume the system is ideal and frictionless. In the absence of non-conservative forces (like air resistance or friction), the total mechanical energy (kinetic + potential) is conserved and remains constant.

Q: Does the potential energy formula $U = \frac{1}{2}kx^2$ hold true here?

Yes! Let's verify: at $x = 0.04\text{ m}$ with $k = 500\text{ N/m}$, the potential energy is $U = \frac{1}{2} \times 500 \times (0.04)^2 = 250 \times 0.0016 = 0.4\text{ J}$. This perfectly matches the given potential energy of $0.4\text{ J}$!

Q: What would happen to the amplitude if the frequency was doubled but the total energy remained the same?

If the frequency is doubled, the spring constant $k$ must increase by a factor of $4$ (since $f \propto \sqrt{k}$). Since $E = \frac{1}{2}kA^2$ is constant, if $k$ increases by $4$, the amplitude squared $A^2$ must decrease by $4$, meaning the amplitude $A$ would be halved.

Q: Why did we not use the position $x = 0.04\text{ m}$ directly in the final amplitude equation?

The amplitude $A$ represents the maximum displacement, whereas $x = 0.04\text{ m}$ is just an instantaneous position. We used the instantaneous energies at $x = 0.04\text{ m}$ to find the total energy $E$, which then allowed us to find the maximum displacement $A$ using $E = \frac{1}{2}kA^2$.

Q: Can amplitude be negative?

No, amplitude is defined as the maximum magnitude of displacement from the equilibrium position, so it is always a positive scalar quantity. The position of the object, however, can be positive or negative ($x = \pm A$).

Question

An object of mass $0.2	ext{ kg}$ executes simple harmonic oscillations along the $X$-axis with a frequency of $\frac{25}{\pi}	ext{ Hz}$. At the position $x = 0.04	ext{ m}$, the object has kinetic energy of $0.5	ext{ J}$ and potential energy $0.4	ext{ J}$. The amplitude of oscillations is ______ $	ext{m}$.

Sigma Percentile · Accepted Answer

### Introduction to Energy in SHM

Simple Harmonic Motion (SHM) is one of the most fundamental and elegant concepts in physics.

At its heart, SHM is a continuous dance between **kinetic energy** and **potential energy**.

As an object oscillates, energy is seamlessly transferred back and forth between these two forms, while the **total mechanical energy** of the system remains absolutely constant, assuming no dissipative forces like friction are at play.

In this problem, we are given a snapshot of an oscillating object of mass $0.2	ext{ kg}$ at a specific position $x = 0.04	ext{ m}$.

At this point, we know both its kinetic energy ($0.5	ext{ J}$) and its potential energy ($0.4	ext{ J}$).

Our mission is to find the **amplitude of oscillation**, which represents the maximum displacement of the object from its equilibrium position.

Let's embark on this journey step-by-step!

---

### Unlocking the Spring Constant ($k$)

Before we can talk about energy and amplitude, we need to understand the physical stiffness of our system, represented by the spring constant $k$.

We are given that the frequency of oscillation is:

$$f = \frac{25}{\pi}	ext{ Hz}$$

The fundamental formula relating frequency, mass, and spring constant is:

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

Let's substitute our known values into this equation:

$$\frac{25}{\pi} = \frac{1}{2\pi}\sqrt{\frac{k}{0.2}}$$

Notice how beautifully the $\pi$ terms on both sides cancel out! Multiplying both sides by $\pi$ gives:

$$25 = \frac{1}{2}\sqrt{\frac{k}{0.2}}$$

Now, let's multiply both sides by $2$ to isolate the square root:

$$50 = \sqrt{\frac{k}{0.2}}$$

To get rid of the square root, we square both sides of the equation:

$$50^2 = \frac{k}{0.2} \implies 2500 = \frac{k}{0.2}$$

Finally, multiplying both sides by $0.2$, we find the spring constant $k$:

$$k = 2500 	imes 0.2 = 500	ext{ N/m}$$

This tells us that our system behaves like a spring with a stiffness of $500	ext{ Newtons per meter}$.

---

### The Power of Energy Conservation

Now that we have the spring constant, let's look at the energy of the system.

According to the **Law of Conservation of Mechanical Energy**, the total mechanical energy $E$ of a simple harmonic oscillator is constant at all points along its path:

$$E = K + U$$

We are given a snapshot of the system at $x = 0.04	ext{ m}$, where:
- Kinetic Energy, $K = 0.5	ext{ J}$
- Potential Energy, $U = 0.4	ext{ J}$

By simply adding these two values, we can find the total mechanical energy of the system:

$$E = 0.5	ext{ J} + 0.4	ext{ J} = 0.9	ext{ J}$$

This total energy of $0.9	ext{ J}$ remains constant throughout the entire motion of the object.

---

### Connecting Total Energy to Amplitude

How does the total energy relate to the amplitude of oscillation?

At the extreme positions ($x = \pm A$), the object momentarily comes to rest, meaning its velocity is zero, and thus its kinetic energy is zero.

At these extreme points, all the mechanical energy is stored purely as potential energy. Therefore, the total energy is equal to the maximum potential energy:

$$E = \frac{1}{2}kA^2$$

We already know both $E$ ($0.9	ext{ J}$) and $k$ ($500	ext{ N/m}$). Let's substitute these values into our equation:

$$0.9 = \frac{1}{2} 	imes 500 	imes A^2$$

Simplifying the right side:

$$0.9 = 250 	imes A^2$$

Now, let's solve for $A^2$:

$$A^2 = \frac{0.9}{250}$$

To make the calculation cleaner, we can multiply the numerator and denominator by $10$:

$$A^2 = \frac{9}{2500}$$

Taking the square root of both sides gives us the amplitude $A$:

$$A = \sqrt{\frac{9}{2500}} = \frac{3}{50} = 0.06	ext{ m}$$

Converting this to centimeters, we get:

$$A = 6	ext{ cm}$$

### Conclusion

The amplitude of oscillation is exactly **$0.06	ext{ m}$** (or $6	ext{ cm}$).

This means the object oscillates between $x = -6	ext{ cm}$ and $x = +6	ext{ cm}$, passing through the equilibrium position with maximum speed, and transferring its energy flawlessly between kinetic and potential states.

Enter Numerical Value:

Visualized Solution

Visualizing the Energy of Simple Harmonic Motion

Connecting Frequency and Spring Constant

Setting up the Frequency Equation

Calculating the Spring Constant $k$

Conservation of Mechanical Energy

Finding the Total Mechanical Energy $E$

Relating Total Energy to Amplitude

Substituting Energy and Spring Constant

Solving for Amplitude $A$

Final Answer and Conclusion

The Sigma Insight: Force and Energy Method in SHM

Introduction to Energy in SHM

Unlocking the Spring Constant ( $k$ )

The Power of Energy Conservation

Connecting Total Energy to Amplitude

Conclusion

Similar Questions

A linear harmonic oscillator of force constant $2 \times 1 0^{6} N/m$ and amplitude $0.01 m$ has a total mechanical energy of $160 J$ . Its (a) maximum potential energy is $100 J$ (b) maximum kinetic energy is $100 J$ (c) maximum potential energy is $160 J$ (d) minimum potential energy is zero

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.

A particle is executing simple harmonic motion (SHM) of amplitude $A$ , along the X-axis, about $x = 0$ . when its potential energy (PE) equals kinetic energy (KE), the position of the particle will be

Starting from the origin, a body oscillates simple harmonically with a period of $2 s$ . After what time will its kinetic energy be $75%$ of the total energy?

Two masses $m_{1}$ and $m_{2}$ are suspended together by a massless spring of spring constant $k$ as shown in the figure. When the masses are in equilibrium, $m_{1}$ is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of $m_{2}$ .

A particle starts executing simple harmonic motion (SHM) of amplitude $a$ and total energy $E$ . At any instant, its kinetic energy is $\frac{3 E}{4}$ , then its displacement $y$ is given by

A body executes simple harmonic motion. The potential energy (PE), the kinetic energy (KE) and total energy (TE) are measured as function of displacement $x$ . Which of the following statements is true?

In a simple harmonic oscillator, at the mean position