Physics Guru

What should be the minimum value of amplitude so that the particle speed can match or exceed the wave speed in sinusoidal progressive wave $y = A sin (ωt − \frac {2\pi}{\lambda}x)$? (A) Particle speed can never exceed wave speed (B) $A_{min} = \lambda$ (C) $A_{min} = \frac {\lambda}{2\pi}$ (D) Particles are not present as the wave travels in vacuum Solution Particle speed = $|\frac {\partial y}{\partial t}| = Aω |cos (ωt − \frac {2\pi}{\lambda}x)|$ So, particle speed is less than or equal to Aω. Wave speed = $\nu.\lambda = \frac {ω}{2\pi}.\lambda$ If particle speed were to match or exceed wave speed, $Aω |cos (ωt − \frac {2\pi}{\lambda} x)| \geq \frac {ω}{2\pi}.\lambda$ $\therefore A \geq \frac{{\frac{\lambda }{{2\pi }}}}{{\left| {cos\left( {\omega t - \frac{{2\pi }}{\lambda }x} \right)} \right|}}$ For A to be min. cosine should be maximum = 1 $A_{min} \geq \frac {\lambda}{2\pi}$ When $A_{min} = \frac {\lambda}{2\pi}$, particle speed can match wave speed periodically and when $A_{min} ...

A stationary man standing on a circular platform that is free to rotate is holding an umbrella starts rotating it anticlockwise. The platform starts rotating clockwise due to conservation of angular momentum. Man also rotates clockwise along with the platform.

Q: As an exercise for wrist, palm and fingers, a person squeezes and releases a rubber ball. Is the work done on the ball positive, negative or zero during (a) compression and (b) during expansion? A: Work done on ball = Increase in elastic potential energy Potential energy increases during compression (think of analogy with spring) and thus the work done is positive. Potential energy decreases during expansion and thus the work done is negative.

Q: In beta decay of a nucleus initially at rest, the recoiling daughter nucleus and the emitted beta particle are sometimes observed not to move along the same straight line. How can this be explained in view of the law of conservation of linear momentum?  A: In beta decay, a nucleus emits a beta particle (electron or positron) and a neutrino (or antineutrino). The neutrino carries away some of the momentum. Since, initial linear momentum is $\vec 0$, conservation of linear momentum requires $\vec 0 = \overrightarrow {{p_R}}  + \overrightarrow {{p_\beta }}  + \overrightarrow {{p_\nu }} $. ​If beta particle and neutrino move in straight line, $\overrightarrow {{p_\nu }}  = k\overrightarrow {{p_\beta }} $ where k is a scalar and therefore $\overrightarrow {{p_R}}  =  - (1 + k)\overrightarrow {{p_\beta }} $ which means the daughter nucleus and the emitted beta particle move along the same straight line.  However, if beta particle and neutrino do not move ...

Let $x>0, y>0, z>0$ satisfy $x+y+z=3$. Prove that, $\frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{3}{2} - \frac{{{{(x - y)}^2} + {{(y - z)}^2} + {{(z - x)}^2}}}{{12}}$ Solution Using Titu Andreescu's lemma, $\frac{{a_1^2}}{{{b_1}}} + \frac{{a_2^2}}{{{b_2}}} + ............. + \frac{{a_n^2}}{{{b_n}}} \ge \frac{{{{({a_1} + {a_2} + ........ + {a_n})}^2}}}{{{b_1} + {b_2} + ....... + {b_n}}}$ for ${b_i} > 0$ We have, $\frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{{{{(x + y + z)}^2}}}{{(1 + x) + (1 + y) + (1 + z)}}$ $ \Rightarrow \frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{{{{(x + y + z)}^2}}}{{3 + (x + y + z)}}$ $ \Rightarrow \frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{3}{2} \because x + y + z = 3$ The expression $\frac{{{{(x - y)}^2} + {{(y - z)}^2} + {{(z - x)}^2}}}{{12}}$ is non-negative. So, if $\frac{{{x^2}}}{...

A man walking at speed v and an old man walking at speed 0.5v enter a circular park through the same gate and walk in opposite directions. If the man completes two rounds and exits through the same gate, how many times will they meet each other? Solution Let the circumference of the circular park be C. Distance covered by the man = 2C Distance covered by the old man = C Relative distance covered = 3C One meeting takes place after one relative round or when C relative distance is covered. So, three meetings would take place after 3C relative distance is covered. This means that just when the man is about to exit from the gate after two rounds, the old man also reaches the gate after one round. The three meetings will be angularly equispaced. In other words there will be a meeting after every 120 degree angle.

Solve the differential equation, $(x - y)\frac{d}{{dx}}\sqrt {\frac{y}{x}}  = x + y + 2\sqrt {xy} ;x > 0,y > 0$ Solution We have, $\left( {\sqrt x  + \sqrt y } \right)\left( {\sqrt x  - \sqrt y } \right)\frac{d}{{dx}}\sqrt {\frac{y}{x}}  = {\left( {\sqrt x  + \sqrt y } \right)^2}$ $ \Rightarrow \frac{d}{{dx}}\sqrt {\frac{y}{x}}  = \frac{{\sqrt x  + \sqrt y }}{{\sqrt x  - \sqrt y }}$ $ \Rightarrow \frac{d}{{dx}}\sqrt {\frac{y}{x}}  = \frac{{1 + \sqrt {\frac{y}{x}} }}{{1 - \sqrt {\frac{y}{x}} }}$ $ \Rightarrow \frac{{dt}}{{dx}} = \frac{{1 + t}}{{1 - t}};t = \sqrt {\frac{y}{x}} $ $ \Rightarrow \frac{{1 - t}}{{1 + t}}dt = dx$ $ \Rightarrow \int {\frac{{2 - (t + 1)}}{{1 + t}}} dt = \int {dx} $ $ \Rightarrow \int {\left( {\frac{2}{{1 + t}} - 1} \right)} dt = \int {dx} $ $ \Rightarrow 2\ln |1 + t| - t = x + C$ $ \Rightarrow 2\ln \left| {1 + \sqrt {\frac{y}{x}} } \right| - \sqrt {\frac{y}{x}}  = x + C$

A particle moves in the xy-plane along the parabolic trajectory $y=x^2$. As the particle approaches the origin, evaluate $L=\mathop {\lim }\limits_{x \to 0} \frac{{{x^2}y({e^{xy}} - 1 - xy)}}{{{{({x^2} + {y^2})}^2}}}$. Solution Substituting $y=x^2$, we have $L=\mathop {\lim }\limits_{x \to 0} \frac{{{x^4}({e^{x^3}} - 1 - x^3)}}{{{{({x^2} + {x^4})}^2}}}$ Or, $L=\mathop {\lim }\limits_{x \to 0} \frac{{{x^4}({e^{x^3}} - 1 - x^3)}}{x^4{{{(1 + {x^2})}^2}}}=\mathop {\lim }\limits_{x \to 0} \frac{{{e^{x^3}} - 1 - x^3}}{{{{(1 + {x^2})}^2}}}$ It may seem like exponential series expansion formula should be used, but it is not required. Direct substitution of x = 0 yields L = 0.

The real solutions to the the septic equation \[{x^7} - 3{x^6} - 8{x^5} + 24{x^4} - 7{x^3} + 21{x^2} - 18x + 54 = 0\] are: (A) 3, 3, -3 (B) -3, -3, 3 (C) No real solutions (D) Seven overall Solution Grouping, $({x^7} - 3{x^6}) + ( - 8{x^5} + 24{x^4}) - (7{x^3} - 21{x^2}) + ( - 18x + 54) = 0$ $ \Rightarrow {x^6}(x - 3) - 8{x^4}(x - 3) - 7{x^2}(x - 3) - 18(x - 3) = 0$ $ \Rightarrow (x - 3)({x^6} - 8{x^4} - 7{x^2} - 18) = 0$ So, x = 3 is one solution. Now, ${x^6} - 8{x^4} - 7{x^2} - 18 = 0$ Let, ${x^2} = t$ $\therefore {t^3} - 8{t^2} - 7t - 18 = 0$ t = 9 satisfies the equation. $\therefore {t^2}(t - 9) + t(t - 9) + 2(t - 9) = 0$ $ \Rightarrow (t - 9)({t^2} + t + 2) = 0$ $\therefore t = 9 = {x^2}$, ${t^2} + t + 2 \ne 0$ since ${t^2} + t + 2 > 0$ $\therefore x =  \pm 3$ $\therefore x = 3,3, - 3$ Hence, (A).

Q: A ball is moving on a horizontal table with some velocity. It stops on its own after moving some distance. Which external force(s) cause(s) change in the linear momentum of the ball (ignore air)? A: The ball loses momentum due to two effects: Friction due to some slipping between the ball and the table. Deformation of the ball and/or the table during rolling, which shifts the line of action of the normal reaction slightly ahead of the vertical through the centre. This gives the contact force a small backward horizontal component (rolling resistance), which opposes the motion and reduces the ball's momentum.

The glowing effect happens because the setting sun hits the waterfall at a certain angle for a few minutes in late February. The atmosphere scatters the sunlight, allowing only the longer red-orange wavelengths to pass through. This remaining red-orange light is what creates the glowing effect.

Evaluate, $I = \int {\frac{{{x^3} + 2x + 1}}{{{x^2} + 1}}} dx$ We have, $I = \int {\frac{{{x^3} + x + x + 1}}{{{x^2} + 1}}} dx$ = $ \int {\frac{{x({x^2} + 1) + x + 1}}{{{x^2} + 1}}} dx$ $ = \int {x + \frac{{x + 1}}{{{x^2} + 1}}dx} $ $ = \int {xdx + \int {\frac{x}{{{x^2} + 1}}} } dx + \int {\frac{{dx}}{{1 + {x^2}}}} $ $ = \frac{{{x^2}}}{2} + \frac{1}{2}\int {\frac{{2x}}{{{x^2} + 1}}} dx + {\tan ^{ - 1}}x + C$ $ = \frac{{{x^2}}}{2} + \frac{1}{2}\ln ({x^2} + 1) + {\tan ^{ - 1}}x + C$

Evaluate, $I = \int\limits_0^\infty {\frac{{dx}}{{{{(1 + {x^2})}^5}}}} $ $Let,x = \tan \theta $ $\therefore dx = {\sec ^2}\theta d\theta $ $\therefore I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sec }^2}\theta }}{{{{(1 + {{\tan }^2}\theta )}^5}}}} d\theta $ $\therefore I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sec }^2}\theta }}{{{{({{\sec }^2}\theta )}^5}}}} d\theta $ $\therefore I = \int\limits_0^{\frac{\pi }{2}} {{{\cos }^8}\theta d\theta } $ $\therefore I = \frac{{7 \cdot 5 \cdot 3 \cdot 1}}{{8 \cdot 6 \cdot 4 \cdot 2}} \times \frac{\pi }{2} = \frac{{7 \cdot 5}}{{8 \cdot 2 \cdot 4 \cdot 2}} \times \frac{{22}}{{7.2}}$ $\therefore I = \frac{{55}}{{128}}$

Q: A bird while flying takes a right turn, where does it get the centripetal force from? A: The centripetal force is provided by the bird’s wings as it changes the direction of its flight. As the bird turns to the right, it must adjust the angle of its wings to produce a component of the force towards the center of the turn.