Kerala Medical Entrance 2007 Questions on Simple harmonic Motion

The following two questions were included from simple harmonic motion, in KEAM (Medical) 2007 question paper:

(1) A simple pendulum has a time period T in vacuum. Its time period when it is completely immersed in a liquid of density one-eighth of the density of material of the bob is

(a) √(7/8) T (b) √(5/8) T (c) √(3/8) T (d) √(8/7) T (e) √(8/5) T

The case of a pendulum immersed in a liquid was briefly discussed in the post dated 16^th August 2006.

The mass of the bob is vρ where ‘v’ is the volume of the bob and ‘ρ’ is the density of the material of the bob. The apparent weight of the bob is v(ρ-σ)g where ‘σ’ is the density of the liquid so that the net value of the downward acceleration of the bob is v(ρ-σ)g /vρ = (1- σ/ρ)g = (1- 1/8)g = (7/8)g.

The period of oscillation of the pendulum suspended in vacuum (or, as usual, in air), is given by

T = 2π√(L/g) where 'L’ is the length and ‘g’ is the acceleration due to gravity.

When the pendulum bob is immersed in the liquid, the net acceleration is (7/8)g and the period becomes

T_liquid = 2π√(8L/7g) = √(8/7) T.

You may see the earlier post on simple pendulum containing other interesting questions, by clicking on the label ‘simple pendulum’ below this post or by making use of the search facility at the top of this page.

(2) A body of mass 20 g connected to a spring of constant K executes SHM with a frequency (5/π) Hz. The value of spring constant is

(a) 4 Nm^–1 (b) 3 Nm^–1 (c) 2 Nm^–1 (d) 5 Nm^–1 (e) 2.5 Nm^–1

This is a standard direct question. The frequency of oscillation of a spring-mass system is given by

N = (1/2π)√(K/m) where K is the spring constant and ‘m’ is the mass.

Therefore, 5/π = (1/2π)√(K/0.02), from which

K = 2 Nm^–1