Question

a stretched string of length 1m and mass 5*10^-4 kg fixed at both ends is under a tension of 20N. if its plucked at point situated at 25cm from one end it would vibrate with a frequency

Answer 1

Hey buddy,

◆ Answer-
f = 400 Hz

◆ Explaination-
# Given-
L = 1 m
l = 25 cm = 0.25 m
m = 5×10^-4 kg
T = 20 N

# Solution-
Mass per unit length -
μ = m/L
μ = 5×10^-4 / 1
μ = 5×10^-4 kg/m

Vibrating frequency is calculated by-
f = (1/2l) √(T/μ)
f = (1 / 2×0.25) √(20 / 5×10^-4)
f = 2 × 200
f = 400 Hz

Therefore, string will vibrate with frequency of 400 Hz.

Hope it helps...

Answer 2

Answer:

200 Hz

Explanation:

Since the wire is plucked at 25 cm... is plucked at 1/4 the length of the string.. it makes 2nd harmonic or first overtone..

==>M=0.0005 kg; L= 1m; T= 20N; linear density,nu= M/L

Thus frequency = 2× v/2l

= 2× 1 / 2L × √[ T / linear density ]

= 2 × ( 1 / 2× 1m) × √[20 / (5×10^-4/1m)]

= 200 Hz

Hope it helped .... Its a JEE Main assured answer

Many students mistake it as the fundamental frequency (but it's not, it's 2nd harmonic) and don't ×2 in the eqn f = v/2l...

All the best for ur upcomingsss.....