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Chapter 14: Problem 26

A small block is attached to an ideal spring and is moving in SHM on a horizontal, friction less surface. The amplitude of the motion is \(0.250 \mathrm{~m}\) and the period is \(3.20 \mathrm{~s}\). What are the speed and acceleration of the block when \(x=0.160 \mathrm{~m} ?\)

Short Answer

Expert verified
The velocity and acceleration of the block when \(x=0.160 \mathrm{~m}\) are roughly 0.981 m/s and -1.236 m/s虏 respectively.

Step by step solution

01

Calculate the angular frequency

Start by calculating the angular frequency. Recall that the angular frequency, \(\omega\), is calculated by the formula \(\omega = \frac{2\pi}{T}\), so \(\omega = \frac{2\pi}{3.20 \mathrm{~s}}\).
02

Calculate the velocity

The velocity in SHM is given by \(v = \omega\sqrt{A^2 - x^2}\). Substitute the values for \(\omega\), \(A\) and \(x\) into the equation, \(v = \frac{2\pi}{3.20 \mathrm{~s}}\sqrt{(0.250 \mathrm{~m})^2 - (0.160 \mathrm{~m})^2}\).
03

Calculate the acceleration

The acceleration in SHM is given by \(a = -\omega^2x\). Substitute the values for \(\omega\) and \(x\) into the equation, \(a = -(\frac{2\pi}{3.20 \mathrm{~s}})^2(0.160 \mathrm{~m})\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Frequency
Angular frequency, denoted by \( \omega \), is a fundamental concept in Simple Harmonic Motion. It describes how quickly a system oscillates in terms of angle per unit of time.
It is defined by the equation \( \omega = \frac{2\pi}{T} \), where \( T \) is the period of the oscillation. The period is the time it takes to complete one full cycle of motion.
This equation highlights that the angular frequency is inversely proportional to the period. Therefore, a shorter period means a higher angular frequency, indicating more rapid oscillations.
  • \( \omega \) is measured in radians per second (rad/s), a unit showing how many radians the system covers in one second.
  • Understanding \( \omega \) helps in predicting the motion characteristics, such as velocity and acceleration, during SHM.
Amplitude
Amplitude, often represented as \( A \), is the maximum extent of a system's oscillation from its equilibrium position. In simple harmonic motion (SHM), amplitude signifies the farthest position the object reaches during its oscillation.
It measures how big the oscillations are, and it's expressed in meters (or whatever units the distance is measured in).
In our context, the amplitude is given as \(0.250 \mathrm{~m}\). This means the block can move 0.250 meters away from its central (equilibrium) position in SHM.
  • The amplitude does not affect the period \( T \) or the angular frequency \( \omega \) of the motion.
  • It is a constant, not affected by the mass of the object or the force restoring it to the equilibrium.
Amplitude tells us the energy stored in the harmonic oscillation; higher amplitude means more energy in the system.
Velocity in SHM
The velocity of an object in simple harmonic motion is determined by several factors, including its position and the angular frequency. The velocity at a specific point, \( x \), can be calculated using the formula \( v = \omega\sqrt{A^2 - x^2} \).
In this formula:
  • \( \omega \) is the angular frequency, determining how fast the system oscillates in terms of rotation.
  • \( A \) is the amplitude, the maximum displacement from equilibrium.
  • \( x \) represents the actual displacement at which you want to find the velocity.
When \( x = 0 \), the velocity is at its maximum, due to maximum kinetic energy and minimum potential energy. Conversely, when \( x = A \), velocity is zero because all the energy is potential.
Acceleration in SHM
In simple harmonic motion, acceleration is directly connected to both the displacement from equilibrium and the square of the angular frequency. It represents how quickly the velocity is changing at any point.
The acceleration \( a \) is calculated using the formula \( a = -\omega^2 x \). This negative sign indicates that acceleration always acts towards the equilibrium point, opposing the displacement.
  • \( \omega^2 \) shows that acceleration is proportional to the square of the angular frequency, making it a rapid rate of change in environments of high \( \omega \).
  • \( x \) is the current displacement from equilibrium, indicating that acceleration is directly proportional to the displacement in magnitude but opposite in direction.
This characteristic ensures the system restores to equilibrium efficiently. When displacement is zero, the acceleration is zero, while maximum acceleration occurs at the maximum displacement.

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Most popular questions from this chapter

An object with mass \(m\) is moving in SHM. It has amplitude \(A_{1}\) and total mechanical energy \(E_{1}\) when the spring has force constant \(k_{1}\) You want to quadruple the total mechanical energy, so \(E_{2}=4 E_{1}\), and halve the amplitude, so \(A_{2}=A_{1} / 2,\) by using a different spring, one with force constant \(k_{2}\). (a) How is \(k_{2}\) related to \(k_{1}\) ? (b) What effect will the change in spring constant and amplitude have on the maximum speed of the moving object?

An object is undergoing SHM with period \(1.200 \mathrm{~s}\) and amplitude \(0.600 \mathrm{~m} .\) At \(t=0\) the object is at \(x=0\) and is moving in the negative \(x\) -direction. How far is the object from the equilibrium position when \(t=0.480 \mathrm{~s} ?\)

DATA Experimenting with pendulums, you attach a light string to the ceiling and attach a small metal sphere to the lower end of the string. When you displace the sphere \(2.00 \mathrm{~m}\) to the left, it nearly touches a vertical wall; with the string taut, you release the sphere from rest. The sphere swings back and forth as a simple pendulum, and you measure its period \(T\). You repeat this act for strings of various lengths \(L\), each time starting the motion with the sphere displaced \(2.00 \mathrm{~m}\) to the left of the vertical position of the string. In each case the sphere's radius is very small compared with \(L\). Your results are given in the table: $$ \begin{array}{l|rrrrrrrr} \boldsymbol{L}(\mathbf{m}) & 12.00 & 10.00 & 8.00 & 6.00 & 5.00 & 4.00 & 3.00 & 2.50 & 2.30 \\ \hline \boldsymbol{T}(\mathbf{s}) & 6.96 & 6.36 & 5.70 & 4.95 & 4.54 & 4.08 & 3.60 & 3.35 & 3.27 \end{array} $$ (a) For the five largest values of \(L,\) graph \(T^{2}\) versus \(L\). Explain why the data points fall close to a straight line. Does the slope of this line have the value you expected? (b) Add the remaining data to your graph. Explain why the data start to deviate from the straight-line fit as \(L\) decreases. To see this effect more clearly, plot \(T / T_{0}\) versus \(L,\) where \(T_{0}=2 \pi \sqrt{L / g}\) and \(g=9.80 \mathrm{~m} / \mathrm{s}^{2} .\) (c) Use your graph of \(T / T_{0}\) versus \(L\) to estimate the angular amplitude of the pendulum (in degrees) for which the equation \(T=2 \pi \sqrt{L / g}\) is in error by \(5 \%\).

A \(2.00 \mathrm{~kg}\) frictionless block attached to an ideal spring with force constant \(315 \mathrm{~N} / \mathrm{m}\) is undergoing simple harmonic motion. When the block has displacement \(+0.200 \mathrm{~m},\) it is moving in the negative \(x\) direction with a speed of \(4.00 \mathrm{~m} / \mathrm{s}\). Find (a) the amplitude of the motion; (b) the block's maximum acceleration; and (c) the maximum force the spring exerts on the block.

DATA You hang various masses \(m\) from the end of a vertical, \(0.250 \mathrm{~kg}\) spring that obeys Hooke's law and is tapered, which means the diameter changes along the length of the spring. since the mass of the spring is not negligible, you must replace \(m\) in the equation \(T=2 \pi \sqrt{m / k}\) with \(m+m_{\text {eff }},\) where \(m_{\text {eff }}\) is the effective mass of the oscillating spring. (See Challenge Problem 14.93.) You vary the mass \(m\) and measure the time for 10 complete oscillations, obtaining these data: $$ \begin{array}{l|lcccc} \boldsymbol{m}(\mathbf{k g}) & 0.100 & 0.200 & 0.300 & 0.400 & 0.500 \\ \hline \text { Time (s) } & 8.7 & 10.5 & 12.2 & 13.9 & 15.1 \end{array} $$ (a) Graph the square of the period \(T\) versus the mass suspended from the spring, and find the straight line of best fit. (b) From the slope of that line, determine the force constant of the spring. (c) From the vertical intercept of the line, determine the spring's effective mass. (d) What fraction is \(m_{\text {eff }}\) of the spring's mass? (e) If a \(0.450 \mathrm{~kg}\) mass oscillates on the end of the spring, find its period, frequency, and angular frequency.
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