A block of mass 5 kg moves along the x-direc | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The force is given by $F = -20x + 10 = -20(x - 0.5)$. Let $X = x - 0.5$. Then $F = -20X$. This represents a Simple Harmonic Motion $(SHM)$ about t

Vedclass · Accepted Answer

$0.5 \ m, -10 \ kg \ m/s$

Vedclass · Answer

(C) The force is given by $F = -20x + 10 = -20(x - 0.5)$. Let $X = x - 0.5$. Then $F = -20X$. This represents a Simple Harmonic Motion $(SHM)$ about the equilibrium position $x_0 = 0.5 \ m$.Comparing with $F = -m\omega^2 X$,we have $m\omega^2 = 20$. With $m = 5 \ kg$,$\omega^2 = 4$,so $\omega = 2 \ rad/s$.The amplitude $A$ is the distance from the equilibrium position to the starting point. At $t = 0$,$x = 1 \ m$,so $A = |1 - 0.5| = 0.5 \ m$.The equation of motion is $X(t) = A \cos(\omega t + \phi)$. At $t = 0$,$X = 0.5$,so $0.5 = 0.5 \cos(\phi) \Rightarrow \phi = 0$.Thus,$x(t) = 0.5 + 0.5 \cos(2t)$.At $t = \pi/4 \ s$,$x = 0.5 + 0.5 \cos(2 \cdot \pi/4) = 0.5 + 0.5 \cos(\pi/2) = 0.5 + 0 = 0.5 \ m$.The velocity is $v = dx/dt = -0.5 \cdot 2 \sin(2t) = -1 \sin(2t)$.At $t = \pi/4 \ s$,$v = -1 \sin(\pi/2) = -1 \ m/s$.Momentum $p = mv = 5 	imes (-1) = -5 \ kg \ m/s$. (Note: Correcting the calculation,the momentum is $-5 \ kg \ m/s$ at $x=0.5 \ m$).

$A$ block of mass $5 \ kg$ moves along the $x$-direction subject to the force $F = (-20x + 10) \ N$,with the value of $x$ in metre. At time $t = 0 \ s$,it is at rest at position $x = 1 \ m$. The position and momentum of the block at $t = (\pi / 4) \ s$ are

Similar Questions

$A$ spring of length $l$ and force constant $k$ is attached to a mass $m$ and oscillates with frequency $f_1$. If the spring is cut into two equal parts and one part is attached to the same mass $m$ to oscillate,its frequency becomes $f_2$. Find the relation between $f_1$ and $f_2$.

$A$ mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude $A$. Its speed as it passes through the equilibrium position is $V$. If extended $2A$ and released,the speed of the mass passing through the equilibrium position will be

The springs in the figure are identical,but the length of the spring in $A$ is three times that in $B$. The ratio of the time periods $T_A/T_B$ is:

$A$ cylindrical block of density $\rho$ is partially immersed in a liquid of density $3\rho$. The plane surface of the block remains parallel to the surface of the liquid. The height of the block is $60\, cm$. The block performs $SHM$ when displaced from its mean position. [Use $g = 9.8\, m/s^2$]

Vedclass Test Series

Exam Paper Generator

Online Exam Module