Chapter 11, Problem 3Q

How could you double the maximum speed of a simple harmonic oscillator (SHO)?

To determine

The way to double the maximum speed of a simple harmonic oscillator (SHO).

Answer to Problem 3Q

Solution:

To double the maximum speed, double the maximum displacement A (amplitude) or double the natural angular frequency of the oscillator $\omega$ .

Explanation of Solution

Introduction:

The solution of a simple harmonic oscillation equation is $A\cos (\omega (t-t_o))$ .

The maximum displacement is A and the maximum speed is $A\omega$ as mentioned before.

The maximum speed is related to the amplitude and the square root of the spring constant and inversely related to the mass.

In SHM, the velocity changes continuously, varying just as the displacement does. At the point where displacement is maximum, the velocity is zero and at zero displacement, the velocity is maximum.

The velocity can be expressed as

  $v=-A\omega \sin (\omega t)$

So, the maximum speed is $A\omega$

  $a=x\text{"- }{\omega }^{2}x$

For the maximum speed to double, the maximum displacement A (amplitude) or the natural angular frequency of the oscillator $\omega$ must be doubled. Doubling $\omega$will depend on the specific oscillator. For instance, the length of a pendulum can be halved or the tension in a string can be doubled or the mass of the spring can be changed.

Conclusion:

For the maximum speed to double, the maximum displacement A (amplitude) or the natural angular frequency of the oscillator $\omega$ must be doubled.