wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The equation for the vibration of a string, fixed at both ends vibrating in its third harmonic, is given by

y=(0.4 cm)sin[(0.314 cm1)x]cos[(600πs1)t].

(a) What is the frequency of vibration ? (b) What are the positions of the nodes? (c) What is the length of the string ? (d) What is the waavelength and the speed of two travelling waves that can interfere to give this vibration ?

Open in App
Solution

The stationary wave equationis given by

y=(0.4 cm)sin[(0.314!cm1)x]cos[(6.00 πs1)t]

(a) ω=600 π

2πf=600 π

f=300 Hz

Wavelength,

λ=2π0.314=(2×3.14)0.314

=20 cm

(b) Therefore Nodes are located at 0, 10 cm, 20 cm, 30 cm.

(c) Length of the string

=3λ2=3×202=30 cm

(d) y=0.4 sin (0.314 x) cos(600 πt)

=0.4 sin{(π10)x}cos(600 πt)

Since, λ and v are the wavelength and velocity of the waves that interfere to give this vibration.

λ=20 cm

v=ωk=600 π(π10)

=6000 cm/sec=60 m/s


flag
Suggest Corrections
thumbs-up
7
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Standing Longitudinal Waves
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon