question_answer

A tuning fork of known frequency 256 Hz makes 5 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was  [AIEEE 2003]

A)            256 + 5 Hz                              

B)            256 + 2Hz

C)            256 ? 2 Hz                              

D)            256 ? 5Hz

Correct Answer: D

Solution :

                   Suppose \[{{n}_{P}}=\] frequency of piano =?  \[({{n}_{P}}\propto \sqrt{T})\]                    \[{{n}_{f}}=\] Frequency of tuning fork \[=256Hz\]                    x = Beat frequency = 5 bps, which is decreasing (5®2) after clanging the tension of piano wire                    Also, tension of piano wire is increasing so \[{{n}_{P}}\downarrow \] Hence nP­? nf = x¯             Wrong            nf ? nP­= x¯               Correct              Þ nP = nf ? x = 256 ? 5       Hz.