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

A 20 cm long string, having a mass of 1.0 g, is fixed at both ends. The tension in the string is 0.5 N. The string is set into vibrations using an external oscillator of frequency 100 Hz. What is the seperation (in cm) between the successive nodes on the string.

A
10 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
5 cm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
20 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
15 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 5 cm
We know that,
Wave speed on a stretched string is given by,
v=Tμ
Where μ is mass per unit length of the string.

From the given data,
v=0.55×103=10 m/s

The wavelength of the wave established
λ=vf=10100=0.1 m or 10 cm
The seperation between two successive nodes
x=λ2=102=5 cm

Thus, option (b) is the correct answer.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon