A tuning fork of frequency \(480 \mathrm{~Hz}\) produces 10 beats/s when sounded with a vibrating sonometer string. What must have been the frequency of the string if a slight increase in tension produces fewer beats per second than before? (A) \(480 \mathrm{~Hz}\) (B) \(490 \mathrm{~Hz}\) (C) \(460 \mathrm{~Hz}\) (D) \(470 \mathrm{~Hz}\)

We know the following variables: - Frequency of the tuning fork, \(f1 = 480\) Hz - Beat frequency, \(bf = 10\) beats/s Now we need to find the frequency of the sonometer string, let's denote it by \(f2\). The formula for beat frequency is: \(bf = |f1 - f2|\)

We know that if the tension is increased, the beat frequency decreases, so we have two possibilities: 1. The tuning fork's frequency is higher than the string's frequency, and increasing the tension makes the string's frequency closer to the fork's frequency. 2. The tuning fork's frequency is lower than the string's frequency, and increasing the tension makes the string's frequency closer to the fork's frequency. Since the beat frequency must decrease when the tension increases, we can rule out the first possibility: if the tuning fork's frequency is higher than the string's frequency, increasing the tension would move string's frequency away from the fork's frequency. That leaves us with the second possibility: the tuning fork's frequency is lower than the string's frequency. Now we'll use the formula for beat frequency: \(bf = |f1 - f2|\) \(10 = |480 - f2|\)

We'll solve the equation for \(f2\): \(f2 = 480 + 10\) or \(f2 = 480 - 10\) Based on our analysis, we determined that the tuning fork's frequency is lower than the string's frequency, so we can use the first equation: \(f2 = 480 + 10 = 490\) Hz Therefore, the frequency of the sonometer string is 490 Hz, which corresponds to option (B).