The amplitude of a particle executing SHM is 3 cm. The displacement at which its kinetic energy will be 25 % more than the potential energy is: cm.

Question

The amplitude of a particle executing SHM is 3 cm. The displacement at which its kinetic energy will be 25 % more than the potential energy is: cm. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

K E=P E+(P E/4) K E=(5/4) P E (1/2) m ω2(A2-x2)=(5/4) × (1/2) m ω2 x2 [ v =ω √ A 2- x 2] A2-x2=(5/4) x2 (9 x2/4)=A2 x=(2/3) A ∴ x=(2/3) × 3 cm x=2 cm

Q. The amplitude of a particle executing SHM is $3 c m$ . The displacement at which its kinetic energy will be $25%$ more than the potential energy is:_______ $c m$ .

$K E = PE + \frac{PE}{4}$
$K E = \frac{5}{4} PE$
$\frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{5}{4} \times \frac{1}{2} m ω^{2} x^{2}$
$[v = ω A^{2} - x^{2}]$
$A^{2} - x^{2} = \frac{5}{4} x^{2}$
$\frac{9 x ^{2}}{4} = A^{2}$
$x = \frac{2}{3} A$
$∴ x = \frac{2}{3} \times 3 c m$
$x = 2 c m$

Q. The amplitude of a particle executing SHM is 3cm. The displacement at which its kinetic energy will be 25% more than the potential energy is:_______ cm.

KE=PE+4PE​ KE=45​PE 21​mω2(A2−x2)=45​×21​mω2x2 [v=ωA2−x2​] A2−x2=45​x2 49x2​=A2 x=32​A ∴x=32​×3cm x=2cm

Q. The amplitude of a particle executing SHM is $3 c m$ . The displacement at which its kinetic energy will be $25%$ more than the potential energy is:_______ $c m$ .

$K E = PE + \frac{PE}{4}$
$K E = \frac{5}{4} PE$
$\frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{5}{4} \times \frac{1}{2} m ω^{2} x^{2}$
$[v = ω A^{2} - x^{2}]$
$A^{2} - x^{2} = \frac{5}{4} x^{2}$
$\frac{9 x ^{2}}{4} = A^{2}$
$x = \frac{2}{3} A$
$∴ x = \frac{2}{3} \times 3 c m$
$x = 2 c m$