Q35 / 117 · NTA Abhyas (JEE Main) 2020 · Physics · Waves and Sound

A 25 cm long string is fixed at both ends. It has a mass of 2.5 g and vibrates in its first overtone under tension. Also given is a closed pipe which is  30 cm long which vibrates in its fundamental frequency. When both of these are sounded simultaneously, 8 beats per second are heard. It is also observed that decreasing the tension in the string also decreases the beat frequency. What is the tension (in N) in the string? [Given, speed of sound in air is 360 ms-1]

Correct answer: 59.29

Solution

μ=2.525=0.1g/cm=10-2kg/m
1st  overtone,

λs=25 cm=0.25 m
⇒fs=1λsTμ

Figure
Pipe in fundamental frequency, λp4=0.3
λp=1.2m
⇒fp=vsλp=3601.2=300 Hz

Figure

∵ By decreasing the tension, beat
frequency is decreased.
∴  fs>fp

⇒fs-fp=8

⇒10.25T10-2-3601.2=8
⇒10×10025T-300=8

⇒T=308×251000

⇒T=59.29 N

Hence, the correct answer is 59.29.

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