Question

An object is undergoing $\mathbf{\text{SHM}}$ with period 0.900 s and amplitude 0.320 m. At $t=$ 0 the object is at $x=$ 0.320 m and is instantaneously at rest. Calculate the time it takes the object to go (a) from $x=$ 0.320 m to $x=$ 0.160 m and (b) from $x=$ 0.160 m to $x=$ 0.

Short answer

Time from 0.320 m to 0.160 m is 0.150 s; from 0.160 m to 0 is 0.075 s.

Step-by-step solution

Understanding the Parameters of SHM

Simple Harmonic Motion (SHM) can be described by the equation $x(t)=A\cos (\omega t+\varphi )$, where $A$ is the amplitude, $\omega$ is the angular frequency, and $\varphi$ is the phase constant. Given that the amplitude $A=0.320\text{ m}$ and the object starts at the maximum amplitude, $\varphi =0$. Thus, $x(t)=0.320\cos (\omega t)$.

Calculating Angular Frequency

The angular frequency $\omega$ is related to the period $T$ by $\omega =\frac{2\pi }{T}$. Given $T=0.900\text{ s}$, we find $\omega =\frac{2\pi }{0.900}\approx 6.98\text{ rad/s}$.

Finding Time from $x=0.320\text{ m}$ to $x=0.160\text{ m}$

To find the time $t_1$ it takes to move from $x=0.320\text{ m}$ to $x=0.160\text{ m}$, we use the equation $0.160=0.320\cos (6.98t_1)$. Solving for $t_1$, we have $\cos (6.98t_1)=0.5$, which implies $6.98t_1={\cos }^{-1}0.5=\frac{\pi }{3}$. Thus, $t_1=\frac{\pi }{3\times 6.98}\approx 0.150\text{ s}$.

Calculating Time from $x=0.160\text{ m}$ to $x=0\text{ m}$

The additional time $t_2$ to go from $x=0.160\text{ m}$ to $x=0$ can be found by solving $0=0.320\cos (6.98(t_1+t_2))$. This empties at $\cos (6.98(t_1+t_2))=0$, which implies $6.98(t_1+t_2)=\frac{\pi }{2}$ or alternatively any suitable increment of $\pi /2$. For the first occurrence merely $6.98t_2=\frac{\pi }{2}-\frac{\pi }{3}=\frac{\pi }{6}$. Solving for $t_2$, we find $t_2=\frac{\pi }{6\times 6.98}\approx 0.075\text{ s}$.

Key concepts

Angular Frequency

Angular frequency, represented by $\omega$, is a crucial concept in understanding Simple Harmonic Motion (SHM). It tells us how fast the oscillations happen. Think of it as the speed at which the object moves through its cycle.

Mathematically, angular frequency is related to the period $T$ of the motion, with the formula:$$\omega =\frac{2\pi }{T}$$Here, $2\pi$ represents a full circle in radians, as every oscillation is essentially a circular movement in motion terms.

In our example, the period $T$ is given as 0.900 seconds. By substituting into the formula, we calculate the angular frequency as approximately 6.98 rad/s.

• The larger the angular frequency, the faster the object oscillates.
• It is measured in radians per second (rad/s).

Understanding $\omega$ helps you predict how quickly or slowly an object in SHM moves across its path.

Amplitude

In the context of Simple Harmonic Motion, amplitude denoted as $A$, represents the maximum displacement of the oscillating object from its central or equilibrium position. Imagine it as the height of a swing; the higher you go, the larger the amplitude.

In our given problem, the amplitude is 0.320 meters. This tells us the furthest distance the object can be from its starting neutral point.

• Amplitude is a constant value in SHM, meaning it does not change over time for the oscillation unless acted on by an external force.
• It defines the intensity or strength of the motion.

Understanding amplitude is crucial because it gives you insights into the energy of the system. Higher amplitudes generally mean more energy is involved in the oscillations.

Phase Constant

The phase constant, represented as $\varphi$, plays a subtle but important role in SHM equations. It determines where in its cycle the object begins its motion.

In our scenario, since the object starts at the maximum amplitude when $t=0$, we understand that the phase constant $\varphi$ is 0. This means the object begins oscillating from the maximum point on its path.

• $\varphi$ can alter the position of the object at $t=0$.
• It's measured in radians, similar to angles in geometry.

Knowing $\varphi$ is vital for accurately predicting the initial condition of the object at any given time, making it easier to map out its movement on the SHM path.

Period of Oscillation

The period of oscillation, denoted by $T$, describes the time required for the object to complete one full cycle of its motion before repeating itself. It's like measuring how long it takes for a pendulum to swing back to its starting point.

For our example, $T=0.900$ seconds indicates that in less than one second, the object completes a full oscillation back to its starting position.

• The period is directly related to the frequency of oscillation.
• A longer period means slower oscillation, while a shorter period signifies faster oscillation.

Understanding the period helps you anticipate when the oscillation will return to a specific point in its cycle, a vital part of predicting and analyzing SHM behavior.