SATHEE: Simple Harmonic Motion 5 Question 8

Simple Harmonic Motion 5 Question 8

13. A particle of mass $m$ is executing oscillation about the origin on the $X$ -axis. Its potential energy is $U (x) = k | x |^{3}$ , where $k$ is a positive constant. If the amplitude of oscillation is $a$ , then its time period $T$ is

$(1998, 2 M)$

(a) proportional to $1 / \sqrt{a}$

(b) independent of $a$

(d) proportional to $a^{3 / 2}$

Passage Based Questions

Passage

When a particle of mass $m$ moves on the $X$ -axis in a potential of the form $V (x) = k x^{2}$ , it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{m / k}$ , as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of $x = 0$ in a way different from $k x^{2}$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass $m$ moving on the $X$ -axis. Its potential energy is $V (x) = α x^{4} (α > 0)$ for

$| x |$ near the origin and becomes a constant equal to $V_{0}$ for $| x | \geq X_{0}$ (see figure).

(2010)

Show Answer

Answer:

Correct Answer: 13. (a)

Solution:

$U (x) = k | x |^{3}$

$∴ [k] = \frac{[U]}{[x^{3}]} = \frac{[M L^{2} T^{- 2}]}{[L^{3}]} = [M L^{- 1} T^{- 2}]$

Now, time period may depend on

$\begin{aligned} T & \propto (mass)^{x} (amplitude)^{y} (k)^{z} \\ [M^{0} L^{0} T] & = [M]^{x} [L]^{y} {[M L^{- 1} T^{- 2}]}^{z} \\ = [M^{x + z} L^{y - z} T^{- 2 z}] \end{aligned}$

Equating the powers, we get

$\begin{aligned} - 2 z = 1 or z = - 1 / 2 \\ y - z = 0 or y = z = - 1 / 2 \end{aligned}$

Hence, $T \propto (amplitude)^{- 1 / 2}) \propto (a)^{- 1 / 2}$

$or T \propto \frac{1}{\sqrt{a}}$

Simple Harmonic Motion 5 Question 8

13. A particle of mass $m$ is executing oscillation about the origin on the $X$ -axis. Its potential energy is $U (x) = k | x |^{3}$ , where $k$ is a positive constant. If the amplitude of oscillation is $a$ , then its time period $T$ is

Passage Based Questions

Passage

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Simple Harmonic Motion 5 Question 8

13. A particle of mass m is executing oscillation about the origin on the X-axis. Its potential energy is U(x)=k|x|3, where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

Passage Based Questions

Passage

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

13. A particle of mass $m$ is executing oscillation about the origin on the $X$ -axis. Its potential energy is $U (x) = k | x |^{3}$ , where $k$ is a positive constant. If the amplitude of oscillation is $a$ , then its time period $T$ is