The energy of a particle executing simple har | vedclass

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

Question

(A) The total energy of a particle in $SHM$ is $E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$. Comparing this with $E = Ax^2 + Bv^2$,we get $A = \frac{k}{2}$

Vedclass · Accepted Answer

Amplitude of $SHM$ is $\sqrt{\frac{E}{A}}$

Vedclass · Answer

(A) The total energy of a particle in $SHM$ is $E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$. Comparing this with $E = Ax^2 + Bv^2$,we get $A = \frac{k}{2}$ and $B = \frac{m}{2}$.$1$. At maximum displacement $(x = A_{amp})$,velocity $v = 0$. Thus,$E = A(A_{amp})^2$,which gives $A_{amp} = \sqrt{\frac{E}{A}}$. (Option $A$ is correct).$2$. At mean position $(x = 0)$,velocity is maximum $(v_{max})$. Thus,$E = B(v_{max})^2$,which gives $v_{max} = \sqrt{\frac{E}{B}}$. (Option $B$ is correct).$3$. Angular frequency $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{2A}{2B}} = \sqrt{\frac{A}{B}}$. The time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{B}{A}}$. (Option $C$ is correct).$4$. Maximum acceleration $a_{max} = \omega^2 A_{amp} = \left(\frac{A}{B}\right) \sqrt{\frac{E}{A}} = \sqrt{\frac{A^2 E}{B^2 A}} = \frac{\sqrt{AE}}{B}$.Wait,checking option $D$: $\frac{\sqrt{EA}}{B}$ is indeed the correct expression for maximum acceleration. Re-evaluating the question: The amplitude in option $A$ was given as $\sqrt{\frac{2E}{A}}$ in the prompt,which is incorrect. Therefore,option $A$ is the incorrect statement.

Column $I$	Column $II$
$(A)$ The object moves on the $x$-axis under a conservative force such that its speed $v = c_1 \sqrt{c_2 - x^2}$,where $c_1, c_2 > 0$.	$(p)$ The object executes simple harmonic motion.
$(B)$ The object moves on the $x$-axis such that its velocity $v = -kx$,where $k > 0$.	$(q)$ The object does not change its direction.
$(C)$ An object is attached to a spring in an elevator accelerating upwards with constant acceleration $a$. The motion is observed from the elevator.	$(r)$ The kinetic energy of the object keeps on decreasing.
$(D)$ The object is projected vertically upwards with speed $2 \sqrt{GM_e / R_e}$.	$(s)$ The object can change its direction only once.

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.

Similar Questions

$A$ clock $S$ is based on the oscillation 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,which of the following is true?

$A$ particle of mass $m$ moves in the potential energy $U(x)$ shown in the figure. The potential energy is given by $U = \frac{1}{2}kx^2$ for $x < 0$ and $U = mgx$ for $x > 0$. The period of the motion when the particle has total energy $E$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module