V=fλ
μT=fλ
5×10−38=100λ
λ=40cm
2λ=20cm
JEE Main 2019 — Physics Waves & Oscillations
A string of length 1m and mass 5g is fixed at both ends. The tension in the string is 8.0N. The string is set into vibration using an external vibrator of frequency 100Hz. The separation between successive nodes on the string is close to
Held on 10 Jan 2019 · Verified 6 Jul 2026.
20.0cm
10.0cm
16.6cm
33.3cm
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