The frequency of oscillation is ((10/π)) (in Hz ) of a particle of mass 0.1 kg which executes SHM along x -axis. The kinetic energy is 0.3 J and potential energy is 0.2 J at position x=0.02 m. The potential energy is zero at mean position. Find the amplitude of oscillations (in meters)

Question

The frequency of oscillation is ((10/π)) (in Hz ) of a particle of mass 0.1 kg which executes SHM along x -axis. The kinetic energy is 0.3 J and potential energy is 0.2 J at position x=0.02 m. The potential energy is zero at mean position. Find the amplitude of oscillations (in meters) (A) (1/2 √10) (B) (1/√10) (C) √10 (D) 2 √10

Tardigrade Admin · Accepted Answer

ω=2 π f=((K/m))1 / 2 K=(2 π f)2 m (1/2) K A2= total energy of oscillation =0.5 J A=√(1.0/K)=(1/2 π f) √(1.0/0.1)=(1/2 √10)

$\frac{1}{2 10}$

$\frac{1}{10}$

$10$

$210$

$ω = 2 π f = (\frac{K}{m})^{1/2}$
$K = (2 π f)^{2} m$
$\frac{1}{2} K A^{2} =$ total energy of oscillation $= 0.5 J$
$A = \frac{1.0}{K} = \frac{1}{2 π f} \frac{1.0}{0.1} = \frac{1}{2 10}$

Q. The frequency of oscillation is (π10​) (in Hz ) of a particle of mass 0.1kg which executes SHM along x -axis. The kinetic energy is 0.3J and potential energy is 0.2J at position x=0.02m. The potential energy is zero at mean position. Find the amplitude of oscillations (in meters)

210​1​

10​1​

10​

210​

ω=2πf=(mK​)1/2 K=(2πf)2m 21​KA2= total energy of oscillation =0.5J A=K1.0​​=2πf1​0.11.0​​=210​1​

$\frac{1}{2 10}$

$\frac{1}{10}$

$10$

$210$

$ω = 2 π f = (\frac{K}{m})^{1/2}$
$K = (2 π f)^{2} m$
$\frac{1}{2} K A^{2} =$ total energy of oscillation $= 0.5 J$
$A = \frac{1.0}{K} = \frac{1}{2 π f} \frac{1.0}{0.1} = \frac{1}{2 10}$