A 16 kg weight hangs in equilibrium from a light string of total length 16 m and linear density μ = 0.01 kg m⁻¹ wrapped round two ideal pulleys separated by 6 m. The horizontal string segment forms three standing-wave loops. If its required vibration frequency is n Hz, enter √n (nearest integer). [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A 16 kg weight hangs in equilibrium from a light string of total length 16 m and linear density μ = 0.01 kg m⁻¹ wrapped round two ideal pulleys separated by 6 m. The horizontal string segment forms three standing-wave loops. If its required vibration frequency is n Hz, enter √n (nearest integer). [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — By symmetry, two supporting string segments hold the 16 kg mass; with g ≈ 10 m s⁻², tension in each is T = mg/2 = 80 N. Step 2 — Transverse speed v = √(T/μ) = √(80/0.01) = 40√5 m s⁻¹. Step 3 — Three loops over length 6 m imply 6 = 3λ/2, so λ = 4 m. Step 4 — Frequency f = v/λ = 10√5 Hz ≈ 22.36 Hz. Step 5 — As source requests √n for numerical frequency n, √22.36 ≈ 4.73 and the nearest integer is 5 (as printed).