Problem 36 - Algebra and Trigonometry

Question

Spring-Mass System The frequency of oscillation of an object suspended on a spring depends on the stiffness \(k\) of the spring (called the spring constant) and the mass \(m\) of the object. If the spring is compressed a distance \(a\) and then allowed to oscillate, its displacement is given by $$ f(t)=a \cos \sqrt{k / m} t $$ (a) A 10 -g mass is suspended from a spring with stiffness \(k=3 .\) If the spring is compressed a distance 5 \(\mathrm{cm}\) and then released, find the equation that describes the oscillation of the spring. (b) Find a general formula for the frequency (in terms of \(k\) and \(m ) .\) (c) How is the frequency affected if the mass is increased? Is the oscillation faster or slower? (d) How is the frequency affected if a stiffer spring is used (larger \(k\) )? Is the oscillation faster or slower?

Step-by-Step Solution

Answer

(a) \( f(t) = 0.05 \cos(17.32t) \) (b) \( f_{\text{frequency}} = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \) (c) Slower (d) Faster

Step 1: Convert Units

Convert the given units to standard units. The mass is given as 10 grams, which needs to be converted to kilograms: \( 10 \text{ g} = 0.01 \text{ kg} \). The compression distance is given as 5 cm, which needs to be converted to meters: \( 5 \text{ cm} = 0.05 \text{ m} \).

Step 2: Substitute Known Values

Substitute the given values into the displacement formula \( f(t) = a \cos\left(\sqrt{\frac{k}{m}}t\right) \). Here, \( a = 0.05 \text{ m} \), \( k = 3 \text{ N/m} \), and \( m = 0.01 \text{ kg} \). This gives us: \[ f(t) = 0.05 \cos\left(\sqrt{\frac{3}{0.01}}t\right) \].

Step 3: Simplify the Expression

Calculate the expression inside the cosine function: \( \sqrt{\frac{3}{0.01}} \). This simplifies to \( \sqrt{300} = 10\sqrt{3} \approx 17.32 \). Thus, the equation becomes \( f(t) = 0.05 \cos(17.32t) \).

Step 4: Formula for Frequency

The frequency of oscillation \( f_{\text{frequency}} \) can be derived from the angular frequency \( \omega = \sqrt{\frac{k}{m}} \). From \( \omega = 2\pi f_{\text{frequency}} \), we can express the frequency as \( f_{\text{frequency}} = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \).

Step 5: Effect of Increasing Mass

Increasing the mass \( m \) will increase the denominator in \( \sqrt{\frac{k}{m}} \), decreasing the value of the square root, and thus decreasing the frequency. Therefore, the oscillation becomes slower.

Step 6: Effect of Increasing Spring Stiffness

Increasing the spring constant \( k \) increases the numerator in \( \sqrt{\frac{k}{m}} \), increasing the value of the square root, and hence increasing the frequency. Therefore, the oscillation becomes faster.

Key Concepts

Frequency of Oscillation

The frequency of oscillation in a spring-mass system tells us how many cycles the system completes in one second. It is a fundamental concept in understanding how the system behaves over time. In physics, the frequency is determined by the formula \[ f_{\text{frequency}} = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \], where:

• \(k\) represents the spring constant, which indicates the stiffness of the spring.
• \(m\) stands for the mass attached to the spring.

To determine this frequency, we can use angular frequency \(\omega = \sqrt{\frac{k}{m}}\). The relationship \(\omega = 2\pi f_{\text{frequency}}\) helps us transform angular frequency into standard frequency by dividing by \(2\pi\).

When the frequency increases, the system oscillates faster, completing more cycles in a given period. Conversely, a lower frequency means it takes longer to complete one cycle. This concept is crucial for predicting how the system will respond to changes in mass or spring stiffness.

Spring Constant

The spring constant, denoted as \(k\), is a measure of the spring's stiffness. It plays a vital role in determining how the system oscillates. In simple terms, the spring constant tells us how much force is needed to compress or stretch the spring by a unit length.

In the formula \(f_{\text{frequency}} = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\), the spring constant \(k\) is in the numerator under the square root. This means that a larger spring constant results in a higher angular frequency and thus a higher frequency of oscillation.

• As \(k\) increases, the spring becomes stiffer.
• This results in the system completing more cycles in less time.

A stiffer spring (higher \(k\)) provides more restoring force that swiftly returns the mass to its equilibrium position, speeding up oscillations. Understanding \(k\) is essential for optimizing systems where specific oscillation speeds are desired.

Mass and Displacement

The interaction between the mass \(m\) and the displacement \(a\) in a spring-mass system is central to how the system oscillates. These factors directly influence the behavior of the system over time.

In the oscillation formula \(f(t) = a \cos(\sqrt{\frac{k}{m}}t)\),

• \(a\) represents the amplitude, or maximum displacement from the equilibrium point.
• \(m\) is the mass of the object attached to the spring, affecting how much it resists acceleration.

An increase in \(m\) affects the denominator \(m\) in \(\sqrt{\frac{k}{m}}\), reducing the oscillation frequency and causing the system to oscillate slower.

Mass impacts the systems by resisting changes in motion, and a larger mass means a slower return to the equilibrium position. Displacement \(a\), however, indicates how far the system can move from equilibrium, affecting the scope of movement but not the frequency of oscillation directly. Understanding these components helps in predicting and adjusting the system's response to external forces.