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

Online Physics Text Books from NCERT

One advice I used to give my students when I had occasion to teach the pre-degree students was to have a copy of NCERT Physics Text Book. Even now the best Physics text books available for 11^th and 12^th classes are those published by NCERT. The authors have taken pains to explain the fundamentals in detail and it is this aspect that makes the books different from nany other books in the market.

NCERT Text Books are now available online. For downloading the books you will find a link “e-Books from NCERT” on the right side of this page or you will find it here.

KEAM 2007 (Medical) Question on Young’s Double Slit

Here is a question on Young’s double slit which appeared in Kerala Medical Entrance 2007 question paper:

In a two slit experiment, with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5×10^–2 m towards the slits, the change in fringe width is 10^–3 m. Then the wave length of light used is (given that distance between the slits is 0.03 mm)

(a) 4000 Ǻ (b) 4500 Ǻ (c) 5000 Ǻ (d) 5500 Ǻ (e) 6000 Ǻ

It is enough to write two equations for the fringe width and solve for the wave length λ. If β is the initial fringe width, we have

β = λD/d where ‘D’ is the initial distance of the screen from the double slit and ‘d’ is the distance between the slits.

When D is decreased, β is decreased. Therefore, when the screen is moved by 5×10^–2 m towards the slits, we have

β –10^–3 = λ(D – 5×10^–2 )/d

Subtracting this equation from the first one, we obtain

10^–3 = λ×(5×10^–2)/d

Since d = 0.03 mm = 0.03×10^–6 m,

λ = (0.03×10^–6)/(5×10^–2) m = 6000×10^–10 m = 6000 Ǻ. [Option (e)].

Mozilla Firefox Browser

It is my belief that the spelling mistakes in my posts are currently reduced thanks to the Mozilla Firefox Browser. This browser senses spelling mistakes and the user can easily spot out the incorrect words. Now I need not resort to time consuming HTML editing for subscripts and superscripts. These are supported by the browser when pasted in to the browser window. I could have saved a lot of time if I had used the Mozilla Firefox Browser earlier!

A Question (MCQ) on Rolling

Here is a question which checks your understanding of basic things in angular motion and simple harmonic motion:

A solid cylinder of mass M and radius R is resting on a horizontal platform which is parallel to the XZ plane. The cylinder can rotate freely about its own axis, which is along the X-direction. The platform is given a linear simple harmonic motion of angular frequency ‘ω’ and amplitude ‘A’ in the Z-direction. If there is no slipping between the cylinder and the platform, the maximum torque acting on the cylinder is

(a) 2MR²Aω² (b) MR²Aω² (c) 2MRAω²

(d) MRAω² (e) MRAω²/2

You may be remembering the expression for maximum acceleration of a simple harmonic motion: a_max = ω²A

[ If you don’t remember the above expression, you may use the simplest form of simple harmonic motion of amplitude ‘A’ and angular frequency ‘ω’ and differentiate it twice:

z = A sin ωt (We write the displacement as ‘z’ since it is in the Z-direction).

a = d²z/dt² = –ω²A sin ωt

Therefore, the maximum acceleration is ω²A].

The maximum angular acceleration (α_max) of the cylinder is given by

α_max = a_max/R = ω²A/R.

The maximum torque (τ_max) on the cylinder is given by

τ_max = α_maxI, where ‘I’ is the moment of inertia of the cylinder about its own axis (which is equal to MR²/2).

Therefore, maximum torque τ_max = (ω²A/R) (MR²/2) = MRAω²/2

You will find an interesting MCQ on rolling at physicsplus: MCQ on Rolling Bodies