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A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 \(\mathrm{cm}\) and a frequency of 0.850 \(\mathrm{Hz}\) . Find (a) the maximum magnitude of the acceleration and of the velocity; (b) the acceleration and speed when the pom- pom's coordinate is \(x=+9.0 \mathrm{cm} ;(\mathrm{c})\) the time required to move from the equilibrium position directly to a point 12.0 \(\mathrm{cm}\) away. (d) Which of the quantities asked for in parts (a), (b), and (c) can be found using the energy approach used in Section \(14.3,\) and which cannot? Explain.

Short Answer

Expert verified
Max velocity: 96.1 cm/s; Max acceleration: 512 cm/s²; Speed at x=9.0 cm: 83.2 cm/s; Time to 12.0 cm: 0.085s. Energy approach finds maximum velocity and speeds at positions.

Step by step solution

01

Understanding Simple Harmonic Motion (SHM) Parameters

In SHM, the position of the object as a function of time is given by \( x(t) = A\cos(2\pi ft + \phi) \). Here \(A = 18.0 \text{ cm} \), \(f = 0.850 \text{ Hz} \). We can find \(\omega = 2\pi f\), which is the angular frequency.
02

Finding Maximum Velocity and Acceleration

Maximum velocity \(v_{max}\) occurs when \(x = 0\) and is given by \( v_{max} = A\omega \). Maximum acceleration \(a_{max}\) is when the displacement is maximum, given by \( a_{max} = A\omega^2 \). Calculate \( \omega = 2\pi \times 0.850 \text{ Hz} \) to find these values.
03

Calculating Acceleration and Speed at x = +9.0 cm

At \( x = 9.0 \text{ cm} \), speed \( v \) can be found using \( v^2 = \omega^2 (A^2 - x^2) \) and acceleration \( a \) is \( a = -\omega^2 x \). Substitute \( x = 9.0 \text{ cm} \) to find these values.
04

Calculating Time to Move 12.0 cm from Equilibrium

To find the time needed to move from 0 to 12.0 cm, use the SHM equation \( x(t) = A\cos(\omega t) \), solve for \( t \) by substituting \( x = 12.0 \text{ cm} \) and rearranging to get \( \cos^{-1}(x/A) = \omega t \). Calculate \( t \) by solving this.
05

Determine Which Quantities Can Be Found Using Energy Approach

The energy approach, based on the idea that \(\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2\), can be used to find maximum velocity, speed at a position, and acceleration indirectly via energy equations. However, time to reach a specific point is not directly solved using energy and requires the equation of motion.

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Key Concepts

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

Maximum Velocity
In Simple Harmonic Motion (SHM), the maximum velocity of an oscillating object occurs as it passes through its equilibrium position, where the displacement is zero. Imagine a cheerleader waving a pom-pom back and forth. As the pom-pom swings through the midpoint of its path, it reaches its highest speed.

The formula for maximum velocity in SHM is given by:
  • \( v_{max} = A\omega \)
Where:
  • \( A \) is the amplitude of the motion
  • \( \omega \) is the angular frequency
To find the angular frequency \( \omega \), use the relation \( \omega = 2\pi f \), where \( f \) is the frequency. Thus, if the amplitude \( A \) is 18 cm and the frequency \( f \) is 0.850 Hz, you can calculate \( \omega \) and consequently the maximum velocity. This concept is key in determining how fast an object in SHM can move.
Maximum Acceleration
Just like velocity, acceleration in SHM also reaches its maximum but when the displacement is also at its maximum, rather than at equilibrium. As the pom-pom reaches the extremities of its path, it experiences its greatest acceleration because it's about to change direction.

The formula for maximum acceleration in SHM is:
  • \( a_{max} = A\omega^2 \)
This is derived from the fact that the restoring force, and hence the acceleration, is greatest when the displacement is largest. By knowing the amplitude \( A \) and angular frequency \( \omega \), you can calculate the maximum acceleration experienced by the object. It's fascinating how in SHM, both velocity and acceleration are closely tied to the amplitude and angular frequency.
SHM Equations
The behaviors of objects in Simple Harmonic Motion can be precisely described using mathematical equations. These equations enable us to predict how the motion evolves over time. An important equation in SHM is the expression for displacement:
  • \( x(t) = A\cos(2\pi ft + \phi) \)
In this equation:
  • \( x(t) \) is the object's position at time \( t \)
  • \( A \) is the amplitude
  • \( f \) is the frequency
  • \( \phi \) is the phase constant, which depends on initial conditions
Additionally, we use equations like:
  • \( v(t) = -A\omega\sin(\omega t + \phi) \)
  • \( a(t) = -A\omega^2\cos(\omega t + \phi) \)
These describe the velocity and acceleration at any given time \( t \). Mastering these equations is essential for understanding and predicting the behavior of oscillatory systems.
Energy Approach
In the analysis of Simple Harmonic Motion, the energy approach is incredibly useful. It is based on the principle of conservation of energy. At any point during the oscillation, the total mechanical energy is constant. The energy is the sum of kinetic and potential energy, given by:
  • \( \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2 \)
Where:
  • \( m \) is the mass of the object
  • \( v \) is the velocity
  • \( k \) is the spring constant
  • \( x \) is the displacement
  • \( A \) is the amplitude
This approach helps us determine properties such as maximum velocity and speed at specific positions without directly using time or displacement-based equations. However, it does not help directly in determining the time taken to reach a specific displacement, which necessitates using the SHM motion equations directly.
Angular Frequency
Angular frequency, denoted \( \omega \), is a fundamental concept in analyzing Simple Harmonic Motion. It connects to how frequently the oscillation cycles through its path in radian measure per second. Unlike regular frequency, which tells us how many cycles occur per second, angular frequency describes the rate of rotation within the cycle itself.

The formula for angular frequency is:
  • \( \omega = 2\pi f \)
In this equation:
  • \( f \) is the frequency of the oscillation
Angular frequency is crucial because it directly influences both maximum velocity and acceleration in SHM, as seen in the equations \( v_{max} = A\omega \) and \( a_{max} = A\omega^2 \). Understanding \( \omega \) gives deep insight into the dynamics of oscillating systems.

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Most popular questions from this chapter

CP SHM of a Floating Object. An object with height \(h,\) mass \(M,\) and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\) (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M,\) and the cross-sectional area \(A\) of the object. You can ignore the damping due to fluid friction (see Section 14.7\()\) .

CALC A Spring with Mass. The preceding problems in this chapter have assumed that the springs had negligible mass. But of course no spring is completely massless. To find the effect of the spring's mass, consider a spring with mass \(M\) , equilibrium length \(L_{0}\), and spring constant \(k\). When stretched or compressed to a length \(L,\) the potential energy is \(\frac{1}{2} k x^{2},\) where \(x=L-L_{0-}\) (a) Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\) . Assume that the speed of points along the length of the spring varies linearly with distance \(I\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. Calculate the kinetic energy of the spring in terms of \(M\) and \(v .\) (Hint: Divide the spring into pieces of length \(d l ;\) find the speed of each piece in terms of \(l, v,\) and \(L ;\) find the mass of each piece in terms of \(d l, M,\) and \(L ;\) and integrate from 0 to \(L\) . The result is not \(\frac{1}{2} M v^{2}\) , since not all of the spring moves with the same speed.) (b) Take the time derivative of the conservation of energy equation, Eq. (14.21), for a mass \(m\) moving on the end of a massless. By comparing your results to Eq. (14.8), which defines \(\omega,\) show that the angular frequency of oscillation is \(\omega=\sqrt{k / m}\) (c) Apply the procedure of part (b) to obtain the angular frequency of oscillation \(\omega\) of the spring considered in part (a). If the effective mass \(M^{\prime}\) of the spring is defined by \(\omega=\sqrt{k / M^{\prime}},\) what is \(M^{\prime}\) in terms of \(M ?\)

An object with mass 0.200 \(\mathrm{kg}\) is acted on by an elastic restoring force with force constant 10.0 \(\mathrm{N} / \mathrm{m}\) . (a) Graph elastic potential energy \(U\) as a function of displacement \(x\) over a range of \(x\) from \(-0.300 \mathrm{m}\) to \(+0.300 \mathrm{m}\) . On your graph, let \(1 \mathrm{cm}=0.05 \mathrm{J}\) vertically and \(1 \mathrm{cm}=0.05 \mathrm{m}\) horizontally. The object is set into oscillation with an initial potential energy of 0.140 \(\mathrm{J}\) and an initial kinetic energy of 0.060 \(\mathrm{J}\) . Answer the following questions by referring to the graph. (b) What is the amplitude of oscillation? (c) What is the potential energy when the displacement is one half the amplitude? (d) At what displacement are the kinetic and potential energies equal? (e) What is the value of the phase angle \(\phi\) if the initial velocity is positive and the initial displacement is negative?

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the block is at \(x=0.280 \mathrm{m},\) the acceleration of the block is \(-5.30 \mathrm{m} / \mathrm{s}^{2} .\) What is the frequency of the motion?

You are watching an object that is moving in SHM. When the object is displaced 0.600 m to the right of its equilibrium position, it has a velocity of 2.20 \(\mathrm{m} / \mathrm{s}\) to the right and an acceleration of 8.40 \(\mathrm{m} / \mathrm{s}^{2}\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

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