Problem Based Learning in Basic Physics - III

Problem Based Learning in Basic Physics - III

 School Science 50(1) Mar2012

A.  K .Mody

V. E.S. College of Arts, Science and Commerce, Sindhi Society

Chembur, Mumbai – 400 071

H. C. Pradhan

HBCSE, TIFR, V. N. PuravMarg, Mankhurd

Mumbai – 400 088

In this article- third in the series of articles we present problems for a problem based learning course from the area of mechanics based on simple harmonic motion, rotational motion, gravitation and waves. We present the learning objectives in this area of basic physics and what each problem tries to achieve with its solution.

In this article, third in the series of Problem Based Learning in Basic Physics, we present a problems onsimple harmonic motion, rotational motion, gravitation and waves.  Methodology and philosophy of selecting these problems are already discussed. (Pradhan 2009, Mody 2011)

To review methodology in brief, we note here that this PBL (Problem Based Learning)starts after students have been introduced to formal structure of Physics.Ideally students would attempt only main problem. If they find it difficult, then depending upon their area of difficulty, right auxiliary problem have to be introduced by teacher who is expected to be a constructivist facilitator. Teacher may choose as per his/her requirement or may construct questions on the spot to guide student to right idea and method.

Problems:

D:  Simple Harmonic Motion

11.A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute SHM with a time period  where m is the mass and ρ is the density of the liquid.    [NCERT EP XI]

Tasks involved in this problem are:

1.      To understand the balance of forces when the log is in equilibrium. 

2.      To find the net force that would try to restore the log back to equilibrium and realize that it is proportional to displacement.

3.      The constant of proportionality should be identified and its relation with time period would lead to the expression for time period.

Many such motions are equivalent and has simple behavior, examples are liquid in a U-tube, magnets in magnetic field, marble in a hemispherical bowl etc… . Treatment here helps understand such motion or any simple deviation that may be possible.

E : Rotational Motion

12.A carpet of mass M , made of inextensible , material is rolled along its length in the form of a cylinder of radius R and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. Calculate the horizontal velocity of the axis of the cylindrical part when its radius reduces to R/2 . [JEE 1990]

Tasks involved in this problem are:

1.      To realize that mass of cylinder here changes as R² and keeps on changing as the carpet unrolls.

2.      The unrolling part of carpet (cylindrical part) has linear as well as rotational motion.

3.      The deciding factor here is the conservation of energy.

13.A uniform disc of mass m and radius R is projected horizontally with velocity v₀ on a rough horizontal floor so that it starts off with a purely sliding motion at  t = 0. After t₀ seconds, it acquires a purely rolling motion as shown.   (i) Calculate the velocity of the centre of mass of the disc at t₀.   (ii) Assuming the coefficient of friction to be μ, calculate t₀. Also calculate the work done by the frictional force as a function of time and the total work done by it over time t much longer than t₀.     [JEE 1997]

Tasks involved in this problem are:

1.      To identify force that causes sliding disc to rotate and roll.

2.      This force causes slowing down (deceleration) of linear motion and torque due to which speeds up rotation of the disc thus to find condition for this.

3.      To find the condition when rolling without slipping begins.

4.      To identify contribution to kinetic energy initially and finally and hence change in kinetic energy.

14.Two smooth spheres made of identical material having masses m and 2m collide with an oblique impact as shown in figure above. The initial velocities of the masses are also shown. The impact force is along the line joining their centres. The coefficient of restitution is 5/9. Calculate the velocities of the masses after the impact and the loss in the kinetic energy.  [Basavraju and Ghosh]

Tasks involved in this problem are:

1.      To understand that collision is inelastic but y-velocity is unaffected.

2.      To apply conservation of momentum and find velocities in x- direction.

3.      Hence work out motion after the collision.

F : Gravitation

15.Earth’s orbit is an ellipse with eccentricity 0.0167. Thus, earth’s distance from the sun and speed as it moves around the sun varies from day to day. This means that the length of the solar day is not constant through the year. Assume that earth’s spin axis is normal to its orbital plane and find out the length of the shortest and the longest day. A day should be taken from noon to noon. Does this explain the variation of length of the day during the year?  [NCERT EP XI]

Tasks involved in this problem are:

1.      To understand how planet moves under Kepler’s law.

2.      To apply Kepler’s law and find angular speeds at apogee and perigee.

3.      To understand mean rotation and length of the day in terms of angular speed.

16.  A satellite is in an elliptic orbit around the earth with apehelion of 6R and perihelion of 2R where R = 6400 km is the radius of the earth. Find the eccentricity of the orbit. Find the velocity of the satellite at apogee and perigee. What should be done if this satellite has to be transferred to a circular orbit of radius 6R ? [G = 6.67×10 ^-11 SI units and M = 6×10²⁴ kg]  [NCERT EP XI]

Tasks involved in this problem are:

To apply conservation of angular momentum in central force (elliptic orbit) to find out how velocity changes as satellite moves in the orbit .

G : Waves

Learning objectives:

1.      Becoming familiar with wave propagation and effect of distance (I µ 1/r² law) and state of motion of observer or source (Doppler effect) on the observed parameters namely: amplitude, intensity, frequency, wavelength, and phase.

2.      Superposition of waves leading to beats, resonance, and response of detector.

3.      To understand mathematical structure dealing with above mentioned points.

17.The displacement of the medium in a sound wave is given by the equation y₁ = A cos (ax + bt), where A, a and b are positive constants. The wave is reflected by an obstacle situated at x = 0. The intensity of the reflected wave is 0.64 times that of the incident wave. ^*

                    i.            What are the wavelength and frequency of the incident wave?  

                  ii.            Write the equation for reflected wave.

                iii.            In the resultant wave formed after reflection, find the maximum and minimum values of the particle speeds in the medium. 

• It illustrates what happens when a wave is reflected, how do the incident and reflected waves interact and phase of the reflected wave

• It introduces students to waves and reflection on a surface.

Tasks involved in this problem are:

a.   To apply principle of super position.

b.   To understand the phase of the reflected wave.

c.   To find intensity using the rule that it is proportional to square of the amplitude.

·         No supporting problem. Students are to be guided to achieve the goal.

·         To recognize what the phase of the reflected wave would be they were asked to tell the difference between the waves reflected from the denser and rarer surface and how it gets incorporated mathematically into equation.

18.Two radio stations broadcast their programmes at the same amplitude A, and at slightly different frequencies ω₁ and ω₂ respectively, where ω₂ – ω₁ = 10³ Hz. A detector receives the signal from the two stations simultaneously. It can only detect signals of intensity ≥ 2A². 

                    i.            Find the time interval between two successive maxima of the intensity of the signal received by the detector.

                  ii.            Find the time for which the detector remains idle in each cycle of intensity of the signal.    [JEE 1993]

·         This problem illustrates effect of beats on detection of waves.

Tasks involved in this problem are:

1.   To represent two waves by proper equation.

2.      To use the Principle of super position for finding resultant amplitude.

3.      To find final intensity proportional to square of the amplitude

4.      To make decision as to when will detector be idle. For this they may have to plot a graph of intensity v/s time.

19.A source of sound is moving along a circular orbit of radius 3 metres with an angular velocity of 10 rad/s. A sound detector located far away from the source is executing linear simple harmonic motion along the line BD with an amplitude BC = CD = 6 m. The frequency of oscillation of the detector is 5/π per second. The source is at the point A when the detector is at the point B. If the source emits a continuous sound wave of frequency 340 Hz, find the maximum and minimum frequencies recorded by the detector.  [JEE 1990]

• This is an example based on Doppler effect. It is interesting as it makes use of symmetric movement of detector due to its simple harmonic motion.

Tasks involved in this problem are:

a.       To apply Doppler effect by incorporating relative motion of source and detector.

b.      To see and recognize the symmetry in motion between source and detector.

c.       To recognize the importance of the fact that source and detector are far away.

Solutions:

D:  Simple Harmonic Motion

11.     Let the log be pressed and let the vertical displacement at the equilibrium position be x₀ . At equilibrium   mg = buoyant force  =Ax₀ρg

When it is displaced  further by x, the buoyant force is A(x + x₀)ρg.

Therefore the net restoring force  =  buoyant force  - weight  =  Axρg   (proportional to x)

Thus  mω² = Aρg and  hence    time period

E : Rotational Motion

12.Mass of the cylindrical part of the carpet is proportional to R² and as centre of mass (axis) of the carpet gets lowered, its gravitational potential energy decreases and gets converted into kinetic energy of the rolling part.

P.E. lost  = =

K. E. gained  = K.E. of axis (CM)   +  rotational K.E. around axis

=   ,        where     

and

                          = 

Comparing the two we get ,

13.    (i) Force of friction opposing motion of the disc is    f_r = μmg

                    which causes acceleration of    a = μg

            Torque acting on the disc due to friction is   τ = μmgR

           and causes angular acceleration                    α  = μmgR/I   where  I = ½mR²

       at time t_o,        v = Rω         v_o – at_o = R (ω_o + αt_o)

                                  v_o = 3μmgt_o       t_o = v_o / 3μmg

                                    ω = αt_o =2v_o/3R      and v = R ω = 2v_o/3

(ii)   W = ΔKE  = ½mv² + ½Iω² - ½mv_o²  = (3/2)mμ²g²t (t – 2t_o)

   At  t = t_o :       W = -(3/2) mμ²g²t_o²  =  -(1/6)mv_o²  : work done by friction (decrease in KE)

14.  denoting velocities for masses 2m and m as u₁ and u₂ before and v₁ and v₂ after collision,

we have   ,  ,  and                                        

 Since impact force is only along x- direction, y components of velocities remain unchanged.

v_1y = u_1y = 4 m/s  and v_2y = u_2y = - 8 m/s

Conservation of momentum in x-direction gives 

and coefficient of restitution                  

  v_1x = -5/3 m/s  and v_2x = 10/3 m/s 

and  θ = tan^-1(v_y/v_x) = -12/5  indicates that

Sphere 1 moves at 113^o to x-axis and sphere 2 moves at 67^o (in 4^th quadrant) to x-axis after collision.

F : Gravitation

15.  Angular momentum is conserved in motion under central force and as per Kepler’s law area velocity is constant,  i.e.  r²ω (at perigee)  =  r²ω (at perigee)  = constant

                                                       

If a is the semi-major axis of earth’s orbit, then r_p= a (1 - e)  and  r_a= a (1 + e)

    since  e = 0.0167

If ω is the mean angular speed of the sun (1^o per day) then it corresponds to mean solar day.

 ω_p = 1.034^o per day and ω_a = 0.967^o  per day

If we consider 361^o rotation of earth for 1 day then at perigee day is 8.1” longer and at apogee it is 7.9” smaller.

This does not explain the actual variation of length of the day during the year.

16.   r_a = a (1 + e)  = 6R     and    r_p = a (1 - e)  = 2R        e = ½

      Angular momentum at perigee   =      Angular momentum at apogee

 mv_pr_p = mv_ar_a        v_a = (1/3)v_p

Energy at perigee   =      Energy at apogee

                        v_a  = 2.28 km/s

                      For r = 6R,   

Thus transfer of orbit requires Δv = 0.95 km/s.  Can be achieved by firing rockets from the satellite.

G : Waves

17.Stationary Waves:

        Incident wave  :y₁ = A cos (ax + bt )

(i)     comparing this with y = a cos (2pnt - 2px/l)

we get,    frequency   n = b/2pand   wavelength   l = 2p/a

(ii)  Reflected wave:       y₂ = 0.8 A cos (ax - bt+ p)

Intensity of the reflected wave is 0.64 times and hence amplitude is 0.8 times that of the incident wave. Sign change due to reflection and additional phase of p due to reflection on denser surface.

(iii) Particle speeds are dy/dt and will be maximum at antinodes and minimum at nodes.

18.Beats:

w₂ – w₁ = 10³Hz   Beat frequency

(i)    T = (w₂ – w₁) ^–1 = 1 ms      Beat period.

(ii)  y₁ = A sin 2pw₁t   and     y₂ = A sin 2pw₂t

Resultant wave  y = y₁+ y₂  = R cos 2p[(w₁ + w₂)/2] t

where  R = 2A sin p(w₂ - w₁) t

          For detector response   I ³ 2A²   or R ³+Ö2 A

This is when    sin p(w₂ - w₁) t  =+­ 1/Ö2

p(w₂ - w₁) t  =p/4, 3p/4, 5p/4, …

\detector remains idle for Dt = T/2 = 1/2(w₂ – w₁) = 0.5 ms

19.Doppler Effect :

Detector is far away from the source and hence radius of the signal becomes insignificant.

Both source frequency of circular motion and detector frequency of SHM are identical.

Source is at A (no longitudinal motion) when detector is at B (at momentary rest) and hence no Doppler shift.

When Detector is at C moving towards D, source is also moving away hence minimum frequency get recorded:      

When Detector is at C moving towards B, source is also moving towards hence maximum frequency get recorded:        

References:

*    Source unknown

1. Basavraju G. and Ghosh D., Mechanics and Thermodynamics, Tata McGraw-Hill, New Delhi (1988)

2. JEE - Joint Entrance Examination for admission to IIT

3. Mody A. K. &Pradhan H. C., ‘Problem Based Learning in Basic Physics – I, School Science 49 (3) Sept 2011

4. Pradhan H.C. &Mody A. K., ‘Constructivism applied to physics teaching for                                     capacity building of undergraduate students’, University News, 47 (21) 4-10, (2009)

5. Physics Textbook for class XI part-I, NCERT New Delhi (2006)

6. NCERT EP XI : Physics Exemplar Problems class XI, NCERT New Delhi (2009)