Simple Harmonic Motion Test - 11

A 1.00×10⁻²⁰ kg particle is vibrating with simple harmonic motion with a period of 1.00×10⁻⁵ s and a maximum speed of 1.00×10³ m/s. The maximum displacement of the particle from the mean position is:

The bob of a simple pendulum of length L is released at time t = 0 from a position of small angular displacement. Its linear displacement at time t is given by :

A particle in SHM is described by the displacement function x(t) = A cos(ωt+ϕ), ω = 2π/T. If the initial (t = 0) position of the particle is 1 cm, its initial velocity is π cm s⁻¹ and its angular frequency is πs⁻¹ , then the amplitude of its motion is:

A body of mass 500 g is attached to a horizontal spring of spring constant 8 π²Nm⁻¹. If the body is pulled to a distance of 10 cm from its mean position, then its frequency of oscillation is :

A weightless spring that has a force constant k oscillates with frequency n when a mass m is suspended from it. The spring is cut into equal halves and a mass 2m is suspended from one part of the spring. The frequency of oscillation will now become:

A piece of wood had dimensions a, b and c. Its relative density is d. It is floating in water such that the side c is vertical. It is now pushed down gently and released. The time period is:

The distance covered by a particle undergoing SHM in one time period is: (amplitude=A)

A smooth narrow tunnel is dug along the chord of the earth. The  time period of vibration of a small ball dropped from one end of the tunnel is

The time period of vibration of a uniform disc of mass 'M' and radius 'R' about an axis perpendicular to the plane of disc and passing from a point at a distance R/2 from the center of the disc is:

Two equations of S.H.M. are y₁ = asin(ωt−α) and y₂ = bcos(ωt−α). The phase difference between the two is: