Question

A string is stretched between two rigid supports separated by 75 cm. There are no resonant frequencies between 420 Hz and 315 Hz. The lowest resonant frequency for the string is

• A. 210 Hz
• B. 180 Hz
• C. 105 Hz
• D. 1050 Hz

Hint:

Physics Tip: For any system with equally spaced resonant frequencies, the fundamental frequency is simply the difference between any two consecutive resonant frequencies: $420 - 315 = 105$ Hz.

Answer 1

The Correct Option is C

Concept: Physics (Oscillations and Waves) - Resonant Frequencies of a Stretched String.

Step 1: Identify the nature of the given frequencies. Since there is no resonant frequency between 315 Hz and 420 Hz, these must be consecutive harmonics. Let 315 Hz be the $n^{th}$ harmonic and 420 Hz be the $(n+1)^{th}$ harmonic.

Step 2: Set up the ratio for consecutive harmonics. The frequency of the $n^{th}$ harmonic is given by $\nu_n = n \cdot \nu_0$, where $\nu_0$ is the fundamental (lowest) frequency. $$ \frac{315}{420} = \frac{n}{n+1} \text{} $$

Step 3: Solve for $n$. $$ 315(n+1) = 420n \text{} $$ $$ 315n + 315 = 420n \text{} $$ $$ 105n = 315 \implies n = 3 \text} $$

Step 4: Calculate the fundamental frequency. The lowest resonant frequency $\nu_0$ is: $$ \nu_0 = \frac{\nu_n}{n} = \frac{315}{3} = 105 \text{ Hz } $$ $$ \therefore \text{The lowest resonant frequency is 105 Hz.} \text{} $$