A string of length L, fixed at its both ends is vibrating in its 1 text st overtone mode. Consider two elements of the string of same small length at positions ℓ1=0.2 L and ℓ2=0.45 L from one end. If K1 and K2 are their respective maximum kinetic energies then

Question

A string of length L, fixed at its both ends is vibrating in its 1 text st overtone mode. Consider two elements of the string of same small length at positions ℓ1=0.2 L and ℓ2=0.45 L from one end. If K1 and K2 are their respective maximum kinetic energies then (A) K1=K2 (B) K1 > K2 (C) K1 < K2 (D) it is not possible to decide the relation

Tardigrade Admin · Accepted Answer

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 antinode. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/1975d4a4daf78bca9f0c6b551d0fecd5-.png" />

Q. A string of length $L$ , fixed at its both ends is vibrating in its $1^{st}$ overtone mode. Consider two elements of the string of same small length at positions $ℓ_{1} = 0.2 L$ and $ℓ_{2} = 0.45 L$ from one end. If $K_{1}$ and $K_{2}$ are their respective maximum kinetic energies then

$K_{1} = K_{2}$

$K_{1} > K_{2}$

$K_{1} < K_{2}$

it is not possible to decide the relation

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 antinode.

Q. A string of length L, fixed at its both ends is vibrating in its 1st overtone mode. Consider two elements of the string of same small length at positions ℓ1​=0.2L and ℓ2​=0.45L from one end. If K1​ and K2​ are their respective maximum kinetic energies then

K1​=K2​

K1​>K2​

K1​<K2​