In SHM, kinetic energy is (1 / 4) text th of the total energy at a displacement equal to (Here A is the amplitude of oscillations.)

Question

In SHM, kinetic energy is (1 / 4) text th of the total energy at a displacement equal to (Here A is the amplitude of oscillations.) (A) (A/2) (B) (A/√2) (C) (√3/2) A (D) (2/√3) A

Tardigrade Admin · Accepted Answer

In SHM, Kinetic energy, K=(1/2) m ω2(A2-x2) Total energy, E=(1/2) m ω2 A2. As K=(E/4) ∴ (1/2) m ω2(A2-x2)=(1/4)((1/2) m ω2 A2) or 4 A2-4 x2=A2 or 4 x2=3 A2 or x=(√3/2) A

Q. In SHM, kinetic energy is $(1/4)^{th}$ of the total energy at a displacement equal to
(Here $A$ is the amplitude of oscillations.)

$\frac{A}{2}$

$\frac{A}{2}$

$\frac{3}{2} A$

$\frac{2}{3} A$

In SHM, Kinetic energy, $K = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
Total energy, $E = \frac{1}{2} m ω^{2} A^{2}$ .
As $K = \frac{E}{4}$
$∴ \frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{1}{4} (\frac{1}{2} m ω^{2} A^{2})$
or $4 A^{2} - 4 x^{2} = A^{2}$ or $4 x^{2} = 3 A^{2}$
or $x = \frac{3}{2} A$

Q. In SHM, kinetic energy is (1/4)th of the total energy at a displacement equal to (Here A is the amplitude of oscillations.)

2A​

2​A​

23​​A

3​2​A

In SHM, Kinetic energy, K=21​mω2(A2−x2) Total energy, E=21​mω2A2. As K=4E​ ∴21​mω2(A2−x2)=41​(21​mω2A2) or 4A2−4x2=A2 or 4x2=3A2 or x=23​​A

Q. In SHM, kinetic energy is $(1/4)^{th}$ of the total energy at a displacement equal to
(Here $A$ is the amplitude of oscillations.)

$\frac{A}{2}$

$\frac{A}{2}$

$\frac{3}{2} A$

$\frac{2}{3} A$

In SHM, Kinetic energy, $K = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
Total energy, $E = \frac{1}{2} m ω^{2} A^{2}$ .
As $K = \frac{E}{4}$
$∴ \frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{1}{4} (\frac{1}{2} m ω^{2} A^{2})$
or $4 A^{2} - 4 x^{2} = A^{2}$ or $4 x^{2} = 3 A^{2}$
or $x = \frac{3}{2} A$