Question

When a mass is attached to a vertical spring, the spring is stretched a distance \(d\). The mass is then pulled down from this position and released It undergoes 50.0 oscillations in \(30.0 \mathrm{~s}\). What was the distance \(d\) ?

Short answer

Answer: The mass stretches the spring by approximately 0.092 meters.

Step-by-step solution

Determine the oscillation frequency

First, we need to find the frequency of the oscillations. We can do this by dividing the total number of oscillations by the total time taken for those oscillations to occur. The given information states that there are 50 oscillations in 30 seconds. Frequency (\(f\)) = (Total oscillations) / (Total time) \(f = \frac{50}{30}\) oscillations per second

Determine the oscillation period

Now that we have the frequency of oscillations, we can find the oscillation period (T), which is the reciprocal of the frequency. Oscillation period (\(T\)) = \(\frac{1}{f}\) \(T = \frac{1}{\frac{50}{30}}\) \(T = \frac{30}{50}\) seconds

Apply the period formula for a spring-mass system

The oscillation period for a spring-mass system is given by the formula: \(T = 2\pi\sqrt{\frac{m}{k}}\) Where \(T\) is the oscillation period, \(m\) is the mass attached to the spring, and \(k\) is the spring constant. Since we are given the oscillation period and want to find the stretch distance \(d\), we will first need to rewrite this formula in terms of \(k\). \(k = \frac{4\pi^2m}{T^2}\)

Apply Hooke's Law

Hooke's Law states that the force exerted by a spring is proportional to its displacement from its equilibrium position. Mathematically, it is given by: \(F = -kd\) Where \(F\) is the force exerted by the spring, \(k\) is the spring constant, and \(d\) is the displacement from its equilibrium position. Since the mass attached to the spring stretches it by a distance \(d\), the force exerted by the spring will be equal to the gravitational force acting on the mass. \(F = mg\) Now, equating both expressions for the force: \(mg = kd\) And, rearranging for \(d\): \(d = \frac{mg}{k}\)

Substitute the expression for \(k\) from Step 3

Substitute the expression for \(k\) from Step 3 into the equation for \(d\) from Step 4: \(d = \frac{mg}{\frac{4\pi^2m}{T^2}}\) Now, cancel out the \(m\) terms: \(d = \frac{gT^2}{4\pi^2}\)

Calculate the distance \(d\)

We now have an expression for \(d\) in terms of known quantities. Plug in the values for \(g\) (the acceleration due to gravity) and \(T\) (the oscillation period) into the equation to find the distance \(d\). \(d = \frac{(9.81\,\text{m/s}^2)(\frac{30}{50})^2}{4\pi^2}\) After performing the calculation: \(d \approx 0.092\) meters The distance by which the mass stretches the spring is approximately 0.092 meters.

Key concepts

Oscillation Frequency

Oscillation frequency is a critical concept in understanding spring-mass systems. It represents the number of oscillations that occur per unit of time. In other words, it tells us how fast the mass is vibrating back and forth on the spring. Given by the symbol 'f', the frequency can be calculated by dividing the total number of oscillations by the total duration for which they occur.

For our problem, with 50 oscillations happening over a span of 30 seconds, the oscillation frequency is expressed as:\[\begin{equation}f = \frac{50}{30}\end{equation}\]Hence, the frequency is approximately 1.67 oscillations per second. This is a fundamental parameter in predicting how the spring-mass system will behave under different conditions.

Hooke's Law

Hooke's Law is a principle of physics that relates the force needed to extend or compress a spring to the distance it is stretched or compressed. Precisely, the Law states that the force 'F' acting on a spring is directly proportional to the displacement 'd' from its equilibrium position. Mathematically, it is represented as:\[\begin{equation}F = -kd\end{equation}\]Here, 'k' stands for the spring constant, and the negative sign signifies that the force applied by the spring is in the opposite direction of the displacement. In our scenario, Hooke's Law enables us to equate the force due to the spring with the gravitational force acting on the mass, allowing us to solve for the stretch distance 'd'.

Spring Constant

The spring constant, denoted by the symbol 'k', is a measure that quantifies a spring's stiffness. A high spring constant implies a stiff spring, while a low spring constant indicates a more flexible one. To determine how much a spring can stretch, we apply this constant within Hooke's Law.In the problem provided, we have to calculate 'k' indirectly using the oscillation period formula for a spring-mass system:\[\begin{equation}k = \frac{4\boldsymbol{\pi}^2m}{T^2}\end{equation}\]The spring constant is essential for predicting the system's behavior and solving for unknowns within the equations of motion for a spring-mass system.

Oscillation Period

The oscillation period, represented by 'T', is the time taken for one complete cycle of the oscillation. This is the reciprocal of the oscillation frequency and is a fundamental property of any oscillatory system. For a spring-mass system, the period formula is:\[\begin{equation}T = 2\boldsymbol{\pi} \sqrt{\frac{m}{k}}\end{equation}\]In our example, we derived the period from the given frequency and used it to reveal the spring constant. Understanding the oscillation period is instrumental in determining how a mass attached to a spring will move over time.