A uniform string fixed at both ends is vibrating in 3rd harmonic and equation y=4(cm)sin[(0.8cm−1)×]cos[(400πs−1)t]The length of the vibrating string is
A
6.75m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
12.45m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
11.8m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
18.7m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C11.8m y=4sin(0.8x)cos(400πt)w=400π=2πff=200Hzv=wk=400π×1000.8m/s3⋅v2l=f⇒200=3×400π×1002×l×0.8⇒l=3×400π4×0.8=300π0.8cm=11.8m