Chapter 5.6, Problem 48E

Guitar String A guitar string is pulled at point P a distance of 3 cm above its rest position. It is then released and vibrates in damped harmonic motion with a frequency of 165 cycles per second. After 2 s, it is observed that the amplitude of the vibration at point P is 0.6 cm.

1. (a) Find the damping constant c.
2. (b) Find an equation that describes the position of point P above its rest position as a function of time. Take t = 0 to be the instant that the string is released.

To determine

To find: The damping constant c for the guitar string which oscillates in damped harmonic motion.

Answer to Problem 48E

The damping constant c for the guitar string is $0.80\text{s}$.

Explanation of Solution

Given:

The amplitude after $2\text{s}$ is $0.6\text{ cm}$, frequency is $165\text{cycles}\text{per}\text{second}$ and the initial displacement of the guitar is $\text{3 cm}$.

Definition used:

The equation for the damped harmonic motion which describes the displacement y of an object at time t is,

$y=k{e}^{-ct}\sin \omega t$ (1)

Calculation:

The general equation for the harmonic motion is,

$y=a\sin \omega t$ (2)

Compare equation (1) and (2),

$a=k{e}^{-ct}$ (3)

Assume $t=0$, when the tuning fork start oscillation. At $t=0$, $e^{-ct}=1$.

Substitute 0 for t, and 1 for $e^{-ct}$ in equation (3).

Calculate the value of a at $t=0$.

$\begin{array}{l}a=k{e}^{-c\cdot 0}\\ a=k\end{array}$

Calculate the value of a at $t=2$.

$\begin{array}{l}a=k{e}^{-c\cdot 2}\\ a=k{e}^{-2c}\end{array}$

Since the value of amplitude after $2\text{s}$ is 0.6. So,

$\frac{a\left(0\right)}{a\left(2\right)}=\frac{3}{0.6}$ (4)

Substitute k for $a\left(0\right)$ and $k^{-2c}$ for $a\left(2\right)$ in equation (4).

$\begin{array}{c}\frac{k}{k{e}^{-2c}}=\frac{3}{0.6}\\ \frac{1}{e^{-2c}}=5\\ e^{-2c}=\frac{1}{5}\end{array}$

Take log on both the sides,

$\begin{array}{c}\ln e^{-2c}=\ln \frac{1}{5}\\ -2c=-1.609\\ c=0.80\end{array}$

Thus the value of damping coefficient for the tuning fork is $0.80$.

To determine

To find: The equation which models the displacement of the guitar string as a function of time.

Answer to Problem 48E

The equation for the displacement of the shock absorber as a function of time

is $f\left(t\right)=3e^{-0.8t}\cos 330\pi t$.

Explanation of Solution

Definition used:

The equation for the damped harmonic motion which describes the maximum displacement y of an object at time t is,

$y=k{e}^{-ct}\cos \omega t$ (5)

Calculation:

Calculate the equation for the displacement of the guitar string from its rest position as follows.

When the guitar string is released, take $t=0$.

The formula to calculate the value of $\omega$ is,

$\omega =2\pi f\left[\therefore f=\frac{1}{p}\right]$

Substitute the value 165 for f in $\omega =2\pi f$.

$\begin{array}{c}\omega =2\pi \times 165\\ =330\pi \end{array}$

The value of $\omega$ is $330\pi$.

Substitute the value 3 for k, $0.8$ for c and $330\pi$ for $\omega$ in equation (5).

$\begin{array}{c}y=3\cdot e^{-0.8t}\cos \left(330\pi t\right)\\ =3\cdot e^{-0.8t}\cos 330\pi t\end{array}$

Hence, the function for the displacement is $y=3e^{-0.8t}\cos 330\pi t$.