Question

'n' waves are produced on a string in 1 second. When the radius of the string is doubled, keeping tension same, the number of waves produced in 1 second for the same harmonic will be

• A. $2n$
• B. $\frac{n}{2}$
• C. $\frac{n}{\sqrt{2}}$
• D. $\sqrt{2}n$

Hint:

Remember the general rule for string components: Frequency is always inversely proportional to the string's radius ($f \propto \frac{1}{r}$). Doubling the radius will always cut the frequency exactly in half, giving $\frac{n}{2}$ instantly without needing to write out the full linear density equations!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A vibrating string initially produces $n$ waves per second (frequency $f_1 = n$). The radius of the string is then doubled ($r_2 = 2r_1$) while keeping the mechanical tension $T$ and length $L$ exactly the same. We need to find the new frequency $f_2$ for the same harmonic wave.

Step 2: Key Formula or Approach:
The fundamental frequency of a vibrating string is inversely proportional to the square root of its linear mass density $\mu$: $$f = \frac{v}{2L} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$ The linear mass density $\mu$ can be rewritten in terms of the material's volumetric density $\rho$ and cross-sectional area $A = \pi r^2$: $$\mu = \frac{\text{Mass}}{\text{Length}} = \rho \cdot A = \rho(\pi r^2)$$ Substituting this into our frequency formula shows how frequency scales with the radius of the string: $$f = \frac{1}{2L}\sqrt{\frac{T}{\rho\pi r^2}} = \frac{1}{2Lr}\sqrt{\frac{T}{\rho\pi}} \implies f \propto \frac{1}{r}$$

Step 3: Detailed Explanation:
Since the frequency is inversely proportional to the radius of the string ($f \propto \frac{1}{r}$), we can set up a simple ratio comparison: $$\frac{f_2}{f_1} = \frac{r_1}{r_2}$$ Substitute our initial frequency $f_1 = n$ and the new radius $r_2 = 2r_1$ into this equation: $$\frac{f_2}{n} = \frac{r_1}{2r_1}$$ The radius terms $r_1$ cancel out perfectly: $$\frac{f_2}{n} = \frac{1}{2} \implies f_2 = \frac{n}{2}$$ This matches option (B).

Step 4: Final Answer:
The number of waves produced in 1 second will be $\frac{n}{2}$, which corresponds to option (B).