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NEET PHYSICSWAVEEasy

Question

The length of the string of a musical instrument is 90 cm90\text{ cm} and has a fundamental frequency of 120 Hz120\text{ Hz}. Where should it be pressed to produce a fundamental frequency of 180 Hz180\text{ Hz}?

A

75 cm75\text{ cm}

B

60 cm60\text{ cm}

C

45 cm45\text{ cm}

D

80 cm80\text{ cm}

Step-by-Step Solution

The fundamental frequency of a stretched string is given by f=v2Lf = \frac{v}{2L}. Since the tension and linear mass density of the string remain constant, the wave speed vv is constant. Therefore, the fundamental frequency is inversely proportional to the vibrating length of the string (f1Lf \propto \frac{1}{L}). We can write the relation as: f1L1=f2L2f_1 L_1 = f_2 L_2. Given: Initial frequency, f1=120 Hzf_1 = 120\text{ Hz} Initial length, L1=90 cmL_1 = 90\text{ cm} Final frequency, f2=180 Hzf_2 = 180\text{ Hz} Let the new length be L2L_2. Substituting the values: 120×90=180×L2120 \times 90 = 180 \times L_2 L2=120×90180=60 cmL_2 = \frac{120 \times 90}{180} = 60\text{ cm}. Therefore, the string should be pressed such that the vibrating length becomes 60 cm60\text{ cm}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWAVElengthstringmusicalinstrumentfundamental

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