Two metallic strings `A` and `B` of different materials are connected in series forming a joint. The strings have similar cross - sectional area. The length of `A` is `l_(A) = 0.3 m` and `B` is `l_(B) = 0.75 m`. One end of the combined string is tied with a support rigidly and the other end is loaded with a block of mass `m` passing over a frictionless pulley. Transverse waves are set up in the combined string using an external source of variable frequency, calculate
a. The lowest frequency for which standing waves are observed such that the joints is a node and
b. The total number of antinodes at this frequency. The densities of ` A` and `B` are `6.3 xx 10^(3) kg // m^(3) and 2.8 xx 10^(3) kg//m^(3)`, respectively.

Explanation

To solve the problem step by step, we will break it down into parts as per the requirements of the question.

### Step 1: Understanding the Problem
We have two strings A and B connected in series, with lengths \( l_A = 0.3 \, \text{m} \) and \( l_B = 0.75 \, \text{m} \). The densities of the strings are given as \( \rho_A = 6.3 \times 10^3 \, \text{kg/m}^3 \) and \( \rho_B = 2.8 \times 10^3 \, \text{kg/m}^3 \). We need to find the lowest frequency for which standing waves are observed with the joint as a node and the total number of antinodes at this frequency.

### Step 2: Relationship Between Frequencies
The frequency of the standing waves in each string can be expressed as:
\[
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