The kinetic energy and the potential energy of a particle executing SHM are equal. The ratio of its displacement and amplitude will be

Question

The kinetic energy and the potential energy of a particle executing SHM are equal. The ratio of its displacement and amplitude will be (A) (1/√2) (B) (√3/2) (C) (1/2) (D) √2

Tardigrade Admin · Accepted Answer

Given KE = PE ⇒ (1/2) m v2=(1/2) k x 2 ⇒ (1/2) m ω2(a2-x2)=(1/2) m ω2 x2 ⇒ a2-x2=x2 ⇒ x2=(a2/2) ⇒ (x/a)=(1/√2)

Q. The kinetic energy and the potential energy of a particle executing SHM are equal. The ratio of its displacement and amplitude will be

$\frac{1}{2}$

$\frac{3}{2}$

$\frac{1}{2}$

$2$

Given $K E = PE \Rightarrow \frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$
$\Rightarrow \frac{1}{2} m ω^{2} (a^{2} - x^{2}) = \frac{1}{2} m ω^{2} x^{2}$
$\Rightarrow a^{2} - x^{2} = x^{2} \Rightarrow x^{2} = \frac{a ^{2}}{2} \Rightarrow \frac{x}{a} = \frac{1}{2}$

Q. The kinetic energy and the potential energy of a particle executing SHM are equal. The ratio of its displacement and amplitude will be

2​1​

23​​

21​

2​

Given KE=PE⇒21​mv2=21​kx2 ⇒21​mω2(a2−x2)=21​mω2x2 ⇒a2−x2=x2⇒x2=2a2​⇒ax​=2​1​

$\frac{1}{2}$

$\frac{3}{2}$

$\frac{1}{2}$

$2$

Given $K E = PE \Rightarrow \frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$
$\Rightarrow \frac{1}{2} m ω^{2} (a^{2} - x^{2}) = \frac{1}{2} m ω^{2} x^{2}$
$\Rightarrow a^{2} - x^{2} = x^{2} \Rightarrow x^{2} = \frac{a ^{2}}{2} \Rightarrow \frac{x}{a} = \frac{1}{2}$