Question

An object is undergoing SHM with period 0.300 s and amplitude 6.00 cm. At \(t =\) 0 the object is instantaneously at rest at \(x =\) 6.00 cm. Calculate the time it takes the object to go from \(x =\) 6.00 cm to \(x = -\)1.50 cm.

Short answer

The object takes approximately 0.087 seconds to move from 6.00 cm to -1.50 cm.

Step-by-step solution

Write the equation for Simple Harmonic Motion

The position \(x\) as a function of time \(t\) for an object in simple harmonic motion (SHM) is given by the formula \(x(t) = A \cos(\omega t + \phi)\), where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant.

Determine the angular frequency

The angular frequency \(\omega\) is related to the period \(T\) by the equation \(\omega = \frac{2\pi}{T}\). Given that the period \(T = 0.300 \text{ s}\), we find \(\omega = \frac{2\pi}{0.300}\).

Calculate the angular frequency

Plug in the period to find \(\omega:\) \[\omega = \frac{2\pi}{0.300} = \frac{2\pi}{0.3} \approx 20.94 \text{ rad/s}.\]

Find the phase constant

At \(t = 0\), the object is at rest at \(x = 6.00 \text{ cm}\) which is the amplitude \(A\). Therefore, the initial phase \(\phi = 0\), since \(\cos(\phi) = 1\) when \(\phi = 0\).

Plug values into SHM equation at new position

We substitute into the SHM equation for \(x = -1.50 \text{ cm}\) which becomes:\[-1.50 = 6.00 \cos(20.94t)\] to solve for \(t\).

Solve for \(t\) using inverse cosine

Rearrange the equation and solve for \(t\):\[\cos(20.94t) = \frac{-1.50}{6.00} = -0.25\]Using inverse cosine gives: \[20.94t = \cos^{-1}(-0.25)\].

Calculate cosine inverse

Compute the inverse cosine: \(\cos^{-1}(-0.25) \approx 1.823\) radians.

Solve for time \(t\)

Solve for \(t\) by dividing:\[t = \frac{1.823}{20.94} \approx 0.087 \text{ seconds} \].

Key concepts

Angular Frequency

Angular frequency in simple harmonic motion (SHM) is a fundamental concept that captures how fast the oscillation occurs. It is denoted by \(\omega\) and calculated using the formula \(\omega = \frac{2\pi}{T}\), where \(T\) is the period of oscillation.
- **Understanding \(\omega = \frac{2\pi}{T}\):** This formula shows that while the period of oscillation represents the time it takes to complete one full cycle, the angular frequency describes how many cycles occur in a unit of time, typically per second. The term "angular" originates because, in SHM, the motion can be compared to the projection of uniform circular motion, where the angle turns over time.
- **Units of Angular Frequency:** The unit for angular frequency is radians per second (rad/s). This unit stems from considering that one complete oscillation corresponds to a rotation of \(2\pi\) radians.
In our given problem, with period \(T = 0.300\) seconds, the angular frequency was calculated as \(\omega \approx 20.94\ \text{rad/s}\). This indicates that the object undergoes approximately 21 angular oscillations every second, reflecting a swift oscillation pace.

Amplitude

Amplitude is a crucial parameter in SHM, as it describes the maximum displacement from the equilibrium position. In our problem, the amplitude \(A\) is given as 6.00 cm.
- **What Amplitude Represents:** It tells us the furthest point from the centre that the object reaches on either side during its motion. It is a measure of the energy stored in the oscillation: a larger amplitude denotes a larger potential for energy storage.
- **Role in Equations:** Amplitude is often part of the formula \(x(t) = A \cos(\omega t + \phi)\), dictating the scale of motion depicted in the cosine function. Physically, amplitude is linked to the peak of the wave in oscillation data—where the object is either at the maximum positive or negative position.
At \(t = 0\) in our exercise, the position \(x = 6.00\) cm indicates that the object is at the maximum displacement, i.e., precisely at its amplitude. This setup elucidates the extent of oscillatory movement and helps in solving subsequent positions over time, as demonstrated in solving for \(x = -1.50\) cm.

Period of Oscillation

The period of oscillation is another key characteristic of simple harmonic motion, defined as the time taken for one complete cycle of motion. In our problem, the period \(T\) is 0.300 seconds.
- **Significance of the Period:** It tells us the duration of time over which the motion repeats itself, making it a crucial factor in determining how quickly the oscillation cycle is completed.
- **Relationship with Frequency:** The frequency, \(f\), which is inversely related to the period (\(f = \frac{1}{T}\)), indicates how many cycles occur per second. The lower the period, the higher the frequency, meaning more cycles per second.
In terms of application, knowing the period allows us to predict when the object will return to particular positions in its motion, aiding in calculations like the one required to find the time when the object goes from \(x = 6.00\) cm to \(x = -1.50\) cm. In essence, it's a measure of how "fast" or "slow" the motion seems.