Advertisement Remove all ads

# A 2⋅00 M-long Rope, Having a Mass of 80 G, is Fixed at One End and is Tied to a Light String at the Other End. the Tension in the String is 256 N. (A) Find the Frequencies of - Physics

Sum

A 2⋅00 m-long rope, having a mass of 80 g, is fixed at one end and is tied to a light string at the other end. The tension in the string is 256 N. (a) Find the frequencies of the fundamental and the first two overtones. (b) Find the wavelength in the fundamental and the first two overtones.

Advertisement Remove all ads

#### Solution

Given:
Length of the long rope (L) = 2.00 m
Mass of the rope = 80 g = 0.08 kg
Tension (T) = 256 N
Linear mass density,
$= \frac{0 . 08}{2} = 0 . 04 \text{ kg/m }$

$Tension, T = 256 N$

$Wave velocity, v = \sqrt{\frac{T}{m}}$

$\Rightarrow v = \sqrt{\left( \frac{256}{0 . 04} \right)} = \frac{160}{2}$

$\Rightarrow v = 80 \text{ m/s }$
For fundamental frequency:

$L = \frac{\lambda}{4}$

$\Rightarrow \lambda = 4L = 4 \times 2 = 8 m$

$\Rightarrow f = \frac{v}{\lambda} = \frac{80}{8} = 10 \text{ Hz }$
(a) The frequency overtones are given below:
$\text{ 1st overtone } = 3f = 30 \text{ Hz }$
$\text{ 2nd overtone } = 5f = 50 \text{ Hz }$
(b) $\lambda = 4l = 4 \times 2 = 8 m$
$\therefore \lambda_1 = \frac{v}{f_1} = \frac{80}{30} = 2 . 67 m$
$\lambda_2 = \frac{v}{f_2} = \frac{80}{50} = 1 . 6 m$
Hence, the wavelengths are 8 m, 2.67 m and 1.6 m, respectively.

Is there an error in this question or solution?
Advertisement Remove all ads

#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 15 Wave Motion and Waves on a String
Q 56 | Page 327
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications

Forgot password?