Question

The equation for the vibration of a string, fixed at both ends vibrating in its third harmonic, is given by

$y=(0.4cm)\sin \left[\right(0.314c{m}^{-1}\left)x\right]\cos \left[\right(600\pi s^{-1}\left)t\right]$

(a) What is the frequency of vibration?

(b) What are the positions of the nodes?

(c) What is the length of the string?

(d) What is the wavelength and the speed of two traveling waves that can interface to give this vibration?

Answer 1

Step 1:Given data :

The equation for the vibration of a string, fixed at both ends vibrating in its third harmonic, is given by

$y=(0.4cm)\sin \left[\right(0.314c{m}^{-1}\left)x\right]\cos \left[\right(600\pi s^{-1}\left)t\right]$

Step 2: Formula used:

Know that the standard equation

$y=A\sin \left(kx\right)\cos \left(\omega t\right)$

Frequency can be obtained using the formula of-

$\omega =2\pi f$

Wavelength can be calculated as-

$\lambda =\frac{2\pi }{k}$

Part (a):

Step 3: Find the frequency of vibration

Compare the given equation with the standard equation:

Therefore,

$$\begin{gathered} \omega =600\pi \\ 2\pi f=600\pi \\ f=300Hz \end{gathered}$$

Therefore the frequency is $f=300Hz$.

Part (b):

Step 4: Find the positions of the nodes:

For nodes compare the given equation with the standard equation.

Therefore,

$\sin \left(0.314x\right)=0$

Then,

$\begin{array}{rcl}0.314x& =& p\pi \\ 0.314x& =& p\times 3.14\\ x& =& \frac{p\times 3.14}{0.314}\\ x& =& 10p\end{array}$

Where p is an integer whose values are 0, 1, 2, and 3 for the third harmonic.

Therefore the nodes are at $0,10cm,20cm,30cm$.

Part (c):

Step 5: Find the length of the string:

Know the length of the string $l=n\frac{\lambda }{2}$ here, the $n$ number of nodes (harmonics), and $\lambda$is the wavelength.

Therefore,

The length

$$\begin{gathered} l=3\frac{20}{2} \\ l=\frac{60}{2} \\ l=30cm \end{gathered}$$

Therefore the length of the string is $l=30cm$.

Part (d):

Step 6: Used the formula of the wavelength $\lambda =\frac{2\pi }{k}$

On comparing general equation with the given equation,

$$\begin{gathered} \lambda =\frac{2\times 3.14}{0.314} \\ \lambda =20cm \end{gathered}$$

Therefore the wavelength is $\lambda =20cm$.

Step 7: Find the speed of two traveling waves that can interface to give this vibration

Simplify the equation,

Therefore

$$\begin{gathered} y=(0.4cm)\sin \left[\right(0.314\left)x\right]\cos \left[\right(600\pi \left)t\right] \\ y=0.4\left\{\left(\frac{\pi }{10}\right)x\right\}\cos \left(600\pi t\right) \end{gathered}$$

As we know,

The velocity of the waves $v=\frac{\omega }{k}$.

$k=\frac{\pi }{10}$

Know that the $\lambda =20cm$

Note that $\lambda$ and $v$are the wavelength and velocity of the waves that interface to give this vibration.

Therefore,

$$\begin{gathered} v=\frac{600\pi }{\left({\displaystyle \frac{\pi }{10}}\right)} \\ v=6000cm/s \\ v=60m/s \end{gathered}$$

Therefore, the velocity is $v=60m/s$.