JEE Advanced
Physics
Waves and Sound
2023
JEE Advanced 2023 (Paper 2)
JEE Advanced Physics Question (2023) — Solution
Question
A string of length 1   m and mass 2 × 10 − 5   kg is under tension T . When the string vibrates, two successive harmonics are found to occur at frequencies 750   Hz and 1000   Hz . The value of tension T is _____ newton.
Step-by-step solution
Given: l = 1   m ,   m = 2 × 10 – 5   kg . As successive harmonics are being given, so it is the case of both ends fixed. Now, f n + 1 - f n = 1000 - 750 ⇒ n + 1 2 l T μ - n 2 l T μ = 250 ⇒ 1 2 l T μ = 250 ⇒ T 2 × 10 - 5 = 250 × 2 × 1 ⇒ T 2 × 10 - 5 = 25 × 10 4 ⇒ T = 50 × 10 - 1 ⇒ T = 5   N
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