Home
Class 11
PHYSICS
A string of length L fixed at its at its...

A string of length `L` fixed at its at its both ends is vibrating in its `1^(st)` overtone mode. Consider two elements of the string of same small length at position `l_1=0.2L` and `l_2=0.45L` from one end. If `K_1` and `K_2` are their respective maximum kinetic energies then

A

`K_(1)=K_(2)`

B

`K_(1)gtK_(2)`

C

`K_(1)ltK_(2)`

D

it is not possible to decide the relation

Text Solution

Verified by Experts

The correct Answer is:
B


It is obvious that particle at 0.2 L will have larger amplitude that particle at 0.45 L, 0.5 L being the node and 0.25 being amplitude.
Promotional Banner

Topper's Solved these Questions

  • WAVES AND ACOUSTICS

    A2Z|Exercise AIIMS Questions|55 Videos
  • WAVES AND ACOUSTICS

    A2Z|Exercise NEET Questions|46 Videos
  • WAVES AND ACOUSTICS

    A2Z|Exercise Doppler Effect|25 Videos
  • VECTORS

    A2Z|Exercise Chapter Test|29 Videos
  • WORK, ENERGY, POWER AND COLLISION

    A2Z|Exercise Chapter Test|29 Videos

Similar Questions

Explore conceptually related problems

A string of length L fixed at both ends vibrates in its first overtone. Then the wavelength will be

A string of length L is fixed at both ends . It is vibrating in its 3rd overtone with maximum amplitude a. The amplitude at a distance L//3 from one end is

A stretched string of length L fixed at both the ends has nodes, where n gt 2 . What is the length of the string in terms of n and lambda ?

If a string fixed at both ends, vibrates in its fourth harmonic, the wavelength is 15 cm. What is the length of the string ?

A string of length 'l' is fixed at both ends. It is vibrating in its 3^(rd) overtone with maximum ampltiude 'a' . The amplitude at a distance (l)/(3) from one end is = sqrt(p)(a)/(2) . Find p .