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Chapter 12: Problem 23

You can double the maximum speed of a simple harmonic oscillator by A. Doubling the amplitude B. Reducing the mass to one-fourth its original value C. Increasing the spring constant to four times its original value D. All of above E. None of above

Short Answer

Expert verified
The correct answer is B and C. Both reducing the mass to one-fourth its original value and increasing the spring constant four times its original value can double the maximum speed of a simple harmonic oscillator. The amplitude doesn't affect the maximum speed and hence neither Option A (Doubling the amplitude) nor Option D (All of above) and E (None of above) are correct.

Step by step solution

01

Analyze Option A

Option A suggests doubling the amplitude. However, the amplitude of a oscillation doesn't affect the maximum speed of a harmonic oscillator. So, this option cannot double the maximum speed.
02

Analyze Option B

Option B suggests reducing the mass to one-fourth. The maximum speed of a SHO is inversely proportional to the square root of the mass, i.e., \(v_{max} 鈭� 1/鈭歮\). So, if the mass is reduced to one-fourth, the speed would be doubled. Hence, this option can double the maximum speed.
03

Analyze Option C

Option C suggests quadrupling the spring constant. The maximum speed of a SHO is directly proportional to the square root of the spring constant, i.e., \(v_{max} 鈭� 鈭歬\). So, if the spring constant is quadrupled, the speed would be doubled. Hence, this option can also double the maximum speed.

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Key Concepts

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

Amplitude
The amplitude of a simple harmonic oscillator is the maximum distance it reaches from its equilibrium or rest position. Think of it as the peak height of a pendulum swing or how far a spring stretches or compresses. Amplitude is a crucial element because it affects the energy stored in the system. A larger amplitude means more potential energy is stored at the peak of the cycle. However, when it comes to maximum speed, amplitude doesn't play a significant role.
  • Amplitude defines: Maximum displacement from equilibrium
  • Relates to energy: Higher amplitude equals more potential energy
  • Does not affect speed directly: The speed is dependent on other factors
Understanding amplitude helps us see the extent of motion but is not directly linked to changes in velocity for this oscillator scenario.
Mass
Mass, as it pertains to a simple harmonic oscillator, is a measure of how much matter the moving object has. Whether it's a weight on a spring or another oscillating system, mass plays a critical role in determining the behavior of the oscillator. A lighter mass means less inertia, allowing quicker acceleration and deceleration. This is why reducing the mass can lead to an increase in speed.
  • Mass is crucial because:
  • The lighter the mass, the faster the acceleration
  • In simple harmonic motion, lighter masses achieve higher maximum speeds
Because maximum speed is inversely proportional to the square root of the mass, halving the weight impacts speed significantly.
Spring Constant
The spring constant, represented as 'k,' defines the stiffness of a spring in a simple harmonic oscillator. A higher spring constant means the spring is stiffer and harder to stretch or compress. If you increase the spring constant, the restoring force, which is the force bringing the system back to equilibrium, becomes stronger.
  • Spring constant reflects: Spring stiffness
  • Higher 'k' values mean greater force needed to displace the spring
  • Affects the speed since stronger resistance results in faster movement back to equilibrium
Since the maximum speed relates directly to the square root of the spring constant, quadrupling 'k' effectively doubles the maximum speed of the system. This concept illustrates the balance between elasticity and speed in harmonic oscillators.

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Most popular questions from this chapter

(a) What are the units of \(\omega\) ? (b) What are the units of \(\omega t\) ?

A simple harmonic oscillator is observed to start its oscillations at the maximum amplitude when \(t=0\). Devise an appropriate solution that is consistent with this initial condition. Repeat when the oscillations start at the equilibrium position when \(t=0\).

Starting with the formula \(x(t)=A \cos (\omega t+\phi)\), (a) use direct differentiation to find the velocity at \(t=0\) and \(t=T / 2\). (b) Verify these results using conservation of energy for this simple harmonic oscillator.

The second-order differential equation that describes simple harmonic motion can be written as follows: $$ \frac{d^{2} x}{d t^{2}}+\omega^{2} x=0 $$ Determine, by differentiating, which of the following functions will satisfy the equation (assume \(A\) and \(\omega\) are constants): (a) \(x(t)=A \cos (\omega t)\), (b) \(x(t)=A \sin (\omega t)\), (c) \(x(t)=A \cos (\omega t)+A \sin (\omega t)\), (d) \(x(t)=A e^{\omega t}\), (e) \(x(t)=\) \(A e^{+\omega t}+A e^{-\omega t}\), (f) \(x(t)=A e^{i \omega t}\), (g) \(x(t)=A e^{+i \omega t}+A e^{-i \omega t}\).

A \(500-\mathrm{g}\) object is attached to a spring with a force constant of \(2.5 \mathrm{~N} / \mathrm{m}\). The object rests on a horizontal surface that has a viscous, oily substance spread evenly on it. The object is pulled \(15 \mathrm{~cm}\) to the right of the equilibrium position and set into harmonic motion. After \(3 \mathrm{~s}\), the amplitude has fallen to \(7 \mathrm{~cm}\) due to frictional losses in the oil. (a) Calculate the natural frequency of the system, (b) the damping constant for the oil, and (c) the frequency of oscillation that will be observed for the motion. (d) How much time will it take before the oscillations have died down to one-tenth of the original amplitude \((1.5 \mathrm{~cm})\) ?
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