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Q.The string of a guitar is 80 cm long and has a fundamental frequency of 112 Hz. If a guitarist wishes to produce a frequency of 160 Hz, where should he press the string?

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2026Subjective· 3mImportance★★★★★
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Since f∝1/Lf \propto 1/L for a fixed string (tension, mass per unit length unchanged), the length needed for the new frequency is found from f1L1=f2L2f_1L_1=f_2L_2.

For a stretched string vibrating in its fundamental mode, the fundamental frequency is:

f=12LTμf = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

Since the guitarist only presses the string (changing its vibrating length LL) without altering the tension TT or the mass per unit length μ\mu, we have f∝1/Lf \propto 1/L, i.e. f1L1=f2L2f_1 L_1 = f_2 L_2 for the same string:

L2=f1L1f2=(112)(80)160=8960160=56 cmL_2 = \frac{f_1 L_1}{f_2} = \frac{(112)(80)}{160} = \frac{8960}{160} = 56\ \text{cm}

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