Question

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At \(t\) = 0 the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude and (b) the phase angle. (c) Write an equation for the position as a function of time.

Short answer

(a) Amplitude is 0.98 m. (b) Phase angle is \( \frac{\pi}{2} \). (c) Position as a function of time: \( x(t) = 0.98 \cos(12.25t + \frac{\pi}{2}) \).

Step-by-step solution

Identify the Equations for Simple Harmonic Motion

For a simple harmonic oscillator, the equation for the position of the block as a function of time is given by: \\[ x(t) = A \cos(\omega t + \phi) \] where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase angle. The angular frequency \( \omega \) can be calculated using the formula \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass of the block.

Calculate the Angular Frequency

Substitute the given values \( k = 300 \text{ N/m} \) and \( m = 2.00 \text{ kg} \) in the angular frequency formula. \[ \omega = \sqrt{\frac{300}{2.00}} = \sqrt{150} = 12.25 \text{ rad/s} \]

Use Initial Conditions to Find Amplitude A

At \( t = 0 \), the block is not stretched, so \( x(0) = 0 \). The velocity of the block at \( t = 0 \) is given as 12 m/s in the negative direction, so \( v(0) = -12 \text{ m/s} \). The velocity at any time is given by \\[ v(t) = -A \omega \sin(\omega t + \phi). \] \Substitute \( v(0) = -12 \text{ m/s} \) and \( \omega = 12.25 \text{ rad/s} \) into the equation: \[ -12 = -A \times 12.25 \times \sin(\phi). \] Solve for \( A \sin(\phi) \): \[ A \sin(\phi) = \frac{12}{12.25} = 0.98. \]

Solve for the Amplitude A

We also know that \( x(0) = A \cos(\phi) = 0 \). Since \( \cos(\phi) = 0 \), \( \phi = \frac{\pi}{2} \). \Then, using the relation \( \sin(\phi) = 1 \), it follows that \\[ A = 0.98. \]

Verify and Write the Equation of Motion

Since the amplitude \( A \) is found to be 0.98, and \( \phi = \frac{\pi}{2} \), write the complete expression for position: \\[ x(t) = 0.98 \cos(12.25t + \frac{\pi}{2}). \] \Now, this equation describes the position of the block as a function of time.

Key concepts

Angular Frequency

In simple harmonic motion, angular frequency (\( \omega \)) is a key concept that defines the speed of oscillation. It represents how fast an object moves through its cycle. In this context, it tells us how frequently the block on the spring completes its oscillations. The formula for angular frequency is:
\[ \omega = \sqrt{\frac{k}{m}} \]where:

• \( k \) is the spring constant.
• \( m \) is the mass of the block.

For the given problem, \( k = 300 \text{ N/m} \) and \( m = 2.00 \text{ kg} \). By substituting these values into the formula, we find:
\[ \omega = \sqrt{\frac{300}{2.00}} = 12.25 \text{ rad/s} \]This result means the block oscillates with an angular frequency of 12.25 radians per second, indicating a brisk oscillation speed.

Amplitude

Amplitude (\( A \)) is the maximum extent of displacement from the equilibrium position in simple harmonic motion. It tells us how far the block moves from its central position during oscillation. In this problem, we determine amplitude using the initial conditions.
At the outset (\( t = 0 \)), the block is moving at -12 m/s, but it is not displaced from its equilibrium (\( x(0) = 0 \)). This velocity condition helps us find:
\[ A \sin(\phi) = \frac{12}{12.25} \approx 0.98 \]Since the position at \( t = 0 \) is 0, the phase angle \( \phi = \frac{\pi}{2} \), making \( \sin(\phi) = 1 \). Therefore, the calculated amplitude:
\[ A = 0.98 \]is the maximum distance the block will travel from its equilibrium position during its motion.

Phase Angle

The phase angle (\( \phi \)) is crucial in simple harmonic motion, as it determines the starting point of the oscillation cycle at time = 0. The angle essentially tweaks the function’s cosine waveform, shifting it horizontally.
The given condition \( x(0) = 0 \) with a negative velocity requires us to solve \( A \cos(\phi) = 0 \). This gives:

• \( \phi = \frac{\pi}{2} \)

Meaning, the block starts its motion as a quarter of a period offset. This offset will be regularly repeated in all cycles, ensuring it accurately represents the initial conditions.

Ideal Spring

An ideal spring is a theoretical model used in physics to study spring dynamics without external effects such as friction or air resistance. It follows Hooke's law impeccably, where the force exerted by the spring is directly proportional to the displacement.

• It has infinite cycles of oscillation without losing energy.
• Its behavior is entirely predictable and linear.

In this problem, the spring's ideal nature ensures that the block’s motion is purely harmonic, allowing the calculations to use the standard simple harmonic motion equations without accounting for energy loss. Thus, it can perfectly oscillate back and forth.

Spring Constant

The spring constant (\( k \)) is a measure of the stiffness of a spring, dictating how much force is needed to extend or compress it by a unit length. In simple harmonic motion, it plays a pivotal role in determining the system oscillations.
For this problem, the spring constant is 300 N/m. This means:

• The spring needs 300 Newtons of force to stretch or compress by one meter.
• It influences both the angular frequency and amplitude of the oscillation.

By having a sizeable spring constant like 300 N/m, the oscillations of the block remain well-controlled and restored to equilibrium efficiently, underscoring the precise rhythmic nature inherent in a higher constant value.