Start Free Mock LoginAlready registered? Login here

Physics - SHM Question with Solution | TestHub

PhysicsSHMCalculation of time periodMedium2 minPYQ_2024

PhysicsMediumsingle choice

The potential energy of a particle of mass 0.1 kg moving along the $X$ -axis, is given by $U = 5 x (x - 4) J$ , where $x$ is in metres. Choose the wrong option.

Options:

Answer:

Solution:

$U = 5 x^{2} - 20 x, \frac{d U}{d x} = 10 x - 20 F = - \frac{d U}{d x} = (- 10 x + 20)$ Let us suppose, $x = (X + 2)$ . Then, $F = - 10 X$ , $F = 0$ at $X = 0$ or $x = 2$ , i.e. $x = 2 m$ is the mean position about which particle is in SHM. $k = 10$ $∴ T = 2 π \frac{m}{k} = 2 π \frac{0.1}{10} = \frac{π}{5} s$ At mean position kinetic energy is maximum and force acting on particle in SHM is variable.

Stream:VITEEESubject:PhysicsTopic:SHMSubtopic:Calculation of time period

⏱ 2mℹ️ Source: PYQ_2024

Doubts & Discussion

Loading discussions...