Question

Suppose an object of mass m is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system), the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}}),$$ where \(k > 0\) is a constant measuring the stiffness of the spring (the larger the value of \(k,\) the stiffer the spring) and \(y\) is positive in the \(u p-\) ward direction. Use equation (2) to answer the following questions. a. Find the second derivative \(\frac{d^{2} y}{d t^{2}}\). b. Verify that \(\frac{d^{2} y}{d t^{2}}=-\frac{k}{m} y\).

Short answer

Question: Verify that the second derivative of the position function \( y = y_{0} \cos (t \sqrt{\frac{k}{m}}) \) is equal to \(-\frac{k}{m} y\), where y is the position of an object, and k and m are constants. Answer: After calculating the second derivative of the position function, we found that \(\frac{d^2y}{dt^2} = -y_{0} \cdot k \cdot \cos (t \sqrt{\frac{k}{m}})\). Then, we compared this expression to \(-\frac{k}{m} y\). Since both expressions were found to be equal, we have verified that the second derivative of the position function is equal to \(-\frac{k}{m} y\).

Step-by-step solution

a) Finding the second derivative

First, we need to find the first derivative of the given position function with respect to time t: $$y = y_{0} \cos (t \sqrt{\frac{k}{m}})$$ Differentiating y with respect to t, we get: $$\frac{dy}{dt} = y_{0} \cdot \frac{d}{dt} \cos (t \sqrt{\frac{k}{m}}) $$ Using chain rule: $$\frac{dy}{dt} = y_{0} \cdot (-\sin(t \sqrt{\frac{k}{m}}))\frac{d}{dt}(t \sqrt{\frac{k}{m}})$$ $$\frac{dy}{dt} = -y_{0} \cdot \left(\sin (t \sqrt{\frac{k}{m}}) \cdot \sqrt{\frac{k}{m}}\right)$$ Now, let's find the second derivative, that is, differentiate \(\frac{dy}{dt}\) with respect to t again: $$\frac{d^2y}{dt^2}= -y_{0}\cdot \frac{d}{dt}\left(\sin (t \sqrt{\frac{k}{m}})\cdot \sqrt{\frac{k}{m}}\right) $$ Using chain rule: $$\frac{d^2y}{dt^2}= -y_{0}\cdot \bigg(\cos (t \sqrt{\frac{k}{m}}) \cdot \left(\sqrt{\frac{k}{m}}\right)^2\bigg)$$ So we have the second derivative: $$\frac{d^2y}{dt^2} = -y_{0} \cdot k \cdot \cos (t \sqrt{\frac{k}{m}})$$

b) Verifying the given relationship

To verify that the second derivative is equal to \(-\frac{k}{m}y\), we need to go back to the original position function and multiply it by -\(\frac{k}{m}\): $$\frac{d^2y}{dt^2} = -\frac{k}{m} y = -\frac{k}{m} \cdot (y_{0} \cos (t \sqrt{\frac{k}{m}}))$$ Comparing this to the second derivative we found in part a: $$-\frac{k}{m} \cdot (y_{0} \cos (t \sqrt{\frac{k}{m}})) = -y_{0} \cdot k \cdot \cos (t \sqrt{\frac{k}{m}})$$ Simplification: $$-\frac{k}{m} \cdot (y_{0} \cos (t \sqrt{\frac{k}{m}})) = -y_{0} \cdot \left(\frac{k}{m} \cdot m \right) \cdot \cos (t \sqrt{\frac{k}{m}})$$ We can now cancel out m in the denominator: $$-\frac{k}{m} \cdot (y_{0} \cos (t \sqrt{\frac{k}{m}})) = -y_{0} \cdot k \cdot \cos (t \sqrt{\frac{k}{m}})$$ Since both expressions are equal, we have verified the relationship: $$\frac{d^2y}{dt^2} = -\frac{k}{m} y$$

Key concepts

Simple Harmonic Motion

Imagine an object bobbing up and down on a spring. This back and forth movement, known as simple harmonic motion (SHM), is a type of periodic oscillation that happens when the force acting on an object is proportional and opposite to its displacement from a midpoint. In the provided exercise, the mass on a spring system is a classic example of SHM. When the mass is displaced, the spring exerts a restoring force that pulls it back towards the equilibrium position, where the net force is zero.

Why is it 'simple'? The force here is straightforward—it's Hooke's Law in action, which states that the force on the mass is proportional to its displacement from equilibrium, with the spring constant k describing that proportionality. The movement is predictable and consistent, epitomizing SHM characteristics. The mathematical description invariably involves a sine or cosine function, leading to solutions like the one provided in the exercise, which exhibits how the displacement changes over time.

Differential Calculus

Differential calculus is the branch of mathematics involving rates of change. It’s all about finding out how one quantity changes in relation to another. For example, if you’re driving and accelerating, differential calculus helps to figure out how your speed is changing at every moment in time, not just the overall change from start to stop.

In the context of our oscillating mass, differential calculus allows us to understand how the position of the mass changes as time progresses. This is done by finding derivatives, which are mathematical expressions that represent the rate of change. The first derivative of a position with respect to time gives us the velocity, while the second derivative, as you'll learn soon, reveals the acceleration of the mass. These developments are crucial for analyzing motion in a wide range of scientific and engineering fields.

Second Derivative

Going deeper into the concept, the second derivative isn't just about rate of change; it's the rate of the rate of change. Think of it as meta-change: it tells us how the velocity of an object is changing, which in the case of motion is linked to acceleration.

In our exercise, the second derivative of the position function with respect to time provided us the object's acceleration due to the spring's force. By differentiating twice, we see how the velocity's change, or the acceleration, is also proportional to the displacement, albeit with an added twist—the second derivative introduces a negative sign, indicating a restoring force. This flip is vital, as it reveals the force acting opposite to displacement, bringing the system back to the equilibrium—the essence of SHM.

Cosine Function

The cosine function, one of the basic trigonometric functions, is central to understanding SHM. It oscillates between -1 and 1 and repeats its pattern every 360 degrees or 2π radians, making it perfect for modeling periodic motion such as the oscillation seen in the spring-mass system.

In reference to our exercise, the cosine function's argument includes the angular frequency of the motion, t×√(k/m). This term embodies how swiftly the mass oscillates—affected by both the spring constant k and mass m. Physically, the cosine's value denotes the displacement of the mass at any given time, capturing every crest and trough of the motion comprehensively. When we differentiate this function with respect to time, as part of solving the differential equation, we see the relationship between the position, velocity, and acceleration in the simplest form possible, signifying the heart of SHM.