An object of mass 0.2 kg executes simple harmonic motion along x-axis with frequency of 25/π Hz. At the position x = 0.04 m, the object has a kinetic energy of 0.5 J and potential energy of 0.4 J. The amplitude of oscillation is equal to

Question

An object of mass 0.2 kg executes simple harmonic motion along x-axis with frequency of 25/π Hz. At the position x = 0.04 m, the object has a kinetic energy of 0.5 J and potential energy of 0.4 J. The amplitude of oscillation is equal to (A) 0.05 m (B) 0.06 m (C) 0.01 m (D) 0.02 m

Tardigrade Admin · Accepted Answer

Total energy, E = (1/2) m ω2 A2 E = (1/2) m ( 2π upsilon)2 A2 : :: : (∵ ω = 2 π upsilon) ∴ A = (1/2 π upsilon) √(2E/m) Putting E = K + U , we get A = (1/2π((25)π))√(2×(0.5+0.4)/0.2)= 0.06 m

Q. An object of mass 0.2 kg executes simple harmonic motion along $x$ -axis with frequency of 25/ $π$ Hz. At the position $x$ = 0.04 m, the object has a kinetic energy of 0.5 J and potential energy of 0.4 J. The amplitude of oscillation is equal to

0.05 m

0.06 m

0.01 m

0.02 m

Total energy, $E = \frac{1}{2} m ω^{2} A^{2}$
$E = \frac{1}{2} m (2 π υ)^{2} A^{2} (∵ ω = 2 π υ)$
$∴ A = \frac{1}{2 π υ} \frac{2 E}{m}$
Putting $E = K + U,$ we get
$A = \frac{1}{2 π ( \frac{25}{π} )} \frac{2 \times ( 0.5 + 0.4 )}{0.2} =$ 0.06 m

Q. An object of mass 0.2 kg executes simple harmonic motion along x-axis with frequency of 25/π Hz. At the position x = 0.04 m, the object has a kinetic energy of 0.5 J and potential energy of 0.4 J. The amplitude of oscillation is equal to