An oscillator of mass M is at rest in its equ | vedclass

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

Question

(B) The oscillator is at its equilibrium position $x = X$. Let $M_n$ be the mass of the system after $n$ collisions.Initially,$M_0 = M = 10$. The part

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}}$

Vedclass · Answer

(B) The oscillator is at its equilibrium position $x = X$. Let $M_n$ be the mass of the system after $n$ collisions.Initially,$M_0 = M = 10$. The particle mass is $m = 5$ and speed is $u = 1$.For the $1^{st}$ collision: $m u = (M + m) v_1 \Rightarrow 5(1) = (10 + 5) v_1 \Rightarrow v_1 = \frac{5}{15} = \frac{1}{3}$.After the $1^{st}$ collision,the system oscillates. It crosses the equilibrium position again. Since it is a harmonic oscillator,it returns to the equilibrium position with speed $v_1$ but in the opposite direction (to the left).For the $2^{nd}$ collision: The particle comes from the right with speed $u=1$. The system has momentum $M_1(-v_1) = 15(-\frac{1}{3}) = -5$. The particle has momentum $m u = 5(1) = 5$. Total momentum $= -5 + 5 = 0$.After the $2^{nd}$ collision,the system is at rest at the equilibrium position. The mass is $M_2 = M + 2m = 10 + 10 = 20$.This pattern repeats: after every even number of collisions,the system comes to rest at the equilibrium position.After $12$ collisions,the system is at rest with mass $M_{12} = M + 12m = 10 + 12(5) = 70$.For the $13^{th}$ collision: $m u = (M_{12} + m) v_{13} \Rightarrow 5(1) = (70 + 5) v_{13} \Rightarrow v_{13} = \frac{5}{75} = \frac{1}{15}$.The total mass after $13$ collisions is $M_{13} = 75$.The energy of the oscillator is $E = \frac{1}{2} M_{13} v_{13}^2 = \frac{1}{2} k A^2$.$\frac{1}{2} (75) (\frac{1}{15})^2 = \frac{1}{2} (1) A^2$.$A^2 = \frac{75}{225} = \frac{1}{3} \Rightarrow A = \frac{1}{\sqrt{3}}$.

Column $I$	Column $II$
$(A)$ Potential energy of a simple pendulum ($y$-axis) as a function of displacement ($x$-axis)	$(p)$ Parabolic curve opening upwards
$(B)$ Displacement ($y$-axis) as a function of time ($x$-axis) for a one-dimensional motion at zero or constant acceleration	$(q)$ Linear graph passing through origin
$(C)$ Range of a projectile ($y$-axis) as a function of its velocity ($x$-axis) when projected at a fixed angle	$(r)$ Linear graph with non-zero intercept
$(D)$ The square of the time period ($y$-axis) of a simple pendulum as a function of its length ($x$-axis)	$(s)$ Parabolic curve opening upwards (starting from origin)

Column-$I$	Column-$II$
$(A)$ Velocity-displacement graph $(\omega \neq 1)$	$(i)$ Straight line
$(B)$ Acceleration-displacement graph	$(ii)$ Sinusoidal
$(C)$ Acceleration-time graph	$(iii)$ Circle
$(D)$ Acceleration-velocity graph $(\omega \neq 1)$	$(iv)$ Ellipse

Similar Questions

$A$ clock $S$ is based on the oscillations of a spring,and a clock $P$ is based on pendulum motion. Both clocks run at the same rate on Earth. On a planet having the same density as Earth but twice the radius,then:

The energy of a particle executing simple harmonic motion is given by $E = Ax^2 + Bv^2$,where $x$ is the displacement from mean position $x = 0$ and $v$ is the velocity of the particle at $x$. Choose the incorrect statement.

Vedclass Test Series

Exam Paper Generator

Online Exam Module