A force acts on a $30$gm particle in such a way that the position of the particle as a function of time is given by $x=3t-4t^2+t^3$, where x is in metres and t is in seconds. The work done during the first $4$ second is?
A wheel of radius $1$ meters rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is:
Two SHMs are represented by the equations
$y1=10sin(3\Omega t+\frac{\Omega }{4})$ and
$y2=5[sin3\Omega t+\sqrt{3}cos 3\Omega t]$. their amplitudes and in the ratio
A simple harmonic oscillator starts from extreme position and covers a displacement half of its amplitude in a time '$t$', the further time taken by it to reach mean position is
The circular motion of a particle whose speed is constant is
Find the distance covered by a particle from time $t=0$ to $t=6\ \sec$. When the particle followsa the movement in straight line according to $y=a\cos \left(\dfrac {\pi}{4}\right)t$:-
The particle executes SHM on a straight line. At two positions its velocity $u$ and $v$ while acceleration, $\alpha$ and $\beta$ respectively $[\beta > \alpha >0]$, the distance between the two positions will be:-
A ball is projected from the ground at angle 0 with the horizontal. After 1 sec it is moving at angle ${ 45 }^{ \circ }$ with the horizontal and after 2s it is moving horizontally. What is the velocity of projection of the ball ? (Take $g=10\quad { ms }^{ -2 }$)
To understand Simple Harmonic Motion as analogous to circular motion,
The actual distance moved along the circle will be the distance moved by the projection on the diameter.