The variation of potential energy of harmonic oscillator is as shown in figure. The spring constant is

Question

The variation of potential energy of harmonic oscillator is as shown in figure. The spring constant is (A) 1 × 102 N / m (B) 150 N / m (C) 0.667 × 102 N / m (D) 3 × 102 N / m

Tardigrade Admin · Accepted Answer

Total potential energy =0.04 J Resting potential energy =0.01 J Maximum kinetic energy =(0.04-0.01) =0.03 J =(1/2) m ω2 a2=(1/2) k a2 0.03=(1/2) × k ×((20/1000))2 k=0.06 × 2500 N / m =150 N / m

Q. The variation of potential energy of harmonic oscillator is as shown in figure. The spring constant is

$1 \times 1 0^{2} N / m$

$150 N / m$

$0.667 \times 1 0^{2} N / m$

$3 \times 1 0^{2} N / m$

Total potential energy $= 0.04 J$
Resting potential energy $= 0.01 J$
Maximum kinetic energy $= (0.04 - 0.01)$
$= 0.03 J = \frac{1}{2} m ω^{2} a^{2} = \frac{1}{2} k a^{2}$
$0.03 = \frac{1}{2} \times k \times (\frac{20}{1000})^{2}$
$k = 0.06 \times 2500 N / m = 150 N / m$

Q. The variation of potential energy of harmonic oscillator is as shown in figure. The spring constant is

1×102N/m

150N/m

0.667×102N/m

3×102N/m

Total potential energy =0.04J Resting potential energy =0.01J Maximum kinetic energy =(0.04−0.01) =0.03J=21​mω2a2=21​ka2 0.03=21​×k×(100020​)2 k=0.06×2500N/m=150N/m

$1 \times 1 0^{2} N / m$

$150 N / m$

$0.667 \times 1 0^{2} N / m$

$3 \times 1 0^{2} N / m$

Total potential energy $= 0.04 J$
Resting potential energy $= 0.01 J$
Maximum kinetic energy $= (0.04 - 0.01)$
$= 0.03 J = \frac{1}{2} m ω^{2} a^{2} = \frac{1}{2} k a^{2}$
$0.03 = \frac{1}{2} \times k \times (\frac{20}{1000})^{2}$
$k = 0.06 \times 2500 N / m = 150 N / m$