Physics - SHM Question with Solution | TestHub

PhysicsSHMKinematicsMedium3 minQB

PhysicsMediummultiple choice

The displacement (in m) of a particle of mass 100 g from its equilibrium position is given by the equation:

Answer:

B, C

Solution:

$y = 0.05 sin 3 π (5 t + 0.4)$ gives $ω = 15 π rad / s$ , $A = 0.05 m$ .

$T = \frac{2 π}{15 π} = \frac{2}{15} = \frac{1}{7.5} s$ : (B) true, (A) false.

$a_{m a x} = ω^{2} A = (15 π)^{2} (0.05) = 11.25 π^{2} m / s^{2}$ : (C) true.

At $x = 0.05 m = A$ the force is maximum, not zero: (D) false.

Answer: B,C

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Kinematics

⏱ 3mℹ️ Source: QB

Loading discussions...