Energy of Simple Harmonic Motion Questions in English

(C) The displacement of a particle in simple harmonic motion starting from the mean position is given by $x = A \sin(\omega t)$.
The velocity of the particle is $v = \frac{dx}{dt} = A\omega \cos(\omega t)$.
The kinetic energy $(K)$ is $K = \frac{1}{2}mv^2 = \frac{1}{2}m A^2 \omega^2 \cos^2(\omega t)$.
The total energy $(E)$ is $E = \frac{1}{2}m A^2 \omega^2$.
The ratio of kinetic energy to total energy is $\frac{K}{E} = \cos^2(\omega t)$.
Given $\frac{K}{E} = \frac{3}{4}$,so $\cos^2(\omega t) = \frac{3}{4}$,which implies $\cos(\omega t) = \frac{\sqrt{3}}{2}$.
This means $\omega t = \frac{\pi}{6}$.
Given the time period $T = 1.5 \ s$,we have $\omega = \frac{2\pi}{T} = \frac{2\pi}{1.5} = \frac{4\pi}{3} \ rad/s$.
Substituting $\omega$ into the equation: $(\frac{4\pi}{3}) t = \frac{\pi}{6}$.
Solving for $t$: $t = \frac{\pi}{6} \times \frac{3}{4\pi} = \frac{3}{24} = \frac{1}{8} \ s$.

(C) The kinetic energy $K$ of a particle executing simple harmonic motion at a displacement $x$ is given by $K = \frac{1}{2} k (A^2 - x^2)$,where $k$ is the force constant and $A$ is the amplitude.
At displacement $x_1 = \frac{A}{4}$,the kinetic energy is $K_1 = \frac{1}{2} k (A^2 - (\frac{A}{4})^2) = \frac{1}{2} k (A^2 - \frac{A^2}{16}) = \frac{1}{2} k (\frac{15A^2}{16})$.
At displacement $x_2 = \frac{A}{2}$,the kinetic energy is $K_2 = \frac{1}{2} k (A^2 - (\frac{A}{2})^2) = \frac{1}{2} k (A^2 - \frac{A^2}{4}) = \frac{1}{2} k (\frac{3A^2}{4}) = \frac{1}{2} k (\frac{12A^2}{16})$.
The ratio of the kinetic energies is $\frac{K_1}{K_2} = \frac{\frac{15A^2}{16}}{\frac{12A^2}{16}} = \frac{15}{12} = \frac{5}{4}$.

(A) Given,position of particle of mass $m = 3 \ kg$ is $x = 0.3 \cos (\omega t)$.
Velocity of particle,$v = \frac{dx}{dt} = \frac{d}{dt} (0.3 \cos \omega t) = -0.3 \omega \sin (\omega t)$.
Kinetic energy $K(t) = \frac{1}{2} m v^2 = \frac{1}{2} m (-0.3 \omega \sin \omega t)^2 = \frac{1}{2} m (0.09 \omega^2 \sin^2 \omega t)$.
At $t_1 = \frac{\pi}{6 \omega}$,$K\left(\frac{\pi}{6 \omega}\right) = \frac{1}{2} m (0.09 \omega^2 \sin^2 \frac{\pi}{6}) = \frac{1}{2} m (0.09 \omega^2) \left(\frac{1}{2}\right)^2 = \frac{1}{2} m (0.09 \omega^2) \left(\frac{1}{4}\right)$.
At $t_2 = \frac{\pi}{3 \omega}$,$K\left(\frac{\pi}{3 \omega}\right) = \frac{1}{2} m (0.09 \omega^2 \sin^2 \frac{\pi}{3}) = \frac{1}{2} m (0.09 \omega^2) \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} m (0.09 \omega^2) \left(\frac{3}{4}\right)$.
Taking the ratio: $\frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)} = \frac{1/4}{3/4} = \frac{1}{3}$.

(B) Given: Mass $m = 1 \ kg$,displacement $y = 6 \sin (100 t + \pi/4) \ cm$.
Comparing with the general equation $y = A \sin (\omega t + \phi)$,we get amplitude $A = 6 \ cm = 0.06 \ m$ and angular frequency $\omega = 100 \ rad/s$.
The maximum kinetic energy $(K_{max})$ of a particle in $SHM$ is given by the formula $K_{max} = \frac{1}{2} m \omega^2 A^2$.
Substituting the values: $K_{max} = \frac{1}{2} \times 1 \times (100)^2 \times (0.06)^2$.
$K_{max} = \frac{1}{2} \times 10000 \times 0.0036$.
$K_{max} = 5000 \times 0.0036 = 18 \ J$.

(A) The kinetic energy $(K.E.)$ and potential energy $(P.E.)$ in $SHM$ are given by:
$K.E. = \frac{1}{2} m \omega^2 (A^2 - y^2)$
$P.E. = \frac{1}{2} m \omega^2 y^2$
Given that $K.E. = 8 \times P.E.$
Substituting the expressions:
$\frac{1}{2} m \omega^2 (A^2 - y^2) = 8 \times \frac{1}{2} m \omega^2 y^2$
$A^2 - y^2 = 8 y^2$
$A^2 = 9 y^2$
$y = \pm \frac{A}{3}$
From the given equation $y = \sqrt{3 \pi} \sin \left(\frac{100}{\pi} t + \frac{\pi}{4}\right)$,the amplitude $A = \sqrt{3 \pi}$.
Therefore,the displacement $y = \frac{\sqrt{3 \pi}}{3} = \frac{\sqrt{3} \sqrt{\pi}}{\sqrt{3} \sqrt{3}} = \sqrt{\frac{\pi}{3}}$.

(C) For a spring-block system undergoing simple harmonic motion, the total mechanical energy is conserved.
The total energy $E$ is given by $E = \frac{1}{2} k A^2$, where $k = 100 \text{ Nm}^{-1}$ and amplitude $A = 8 \text{ cm} = 0.08 \text{ m}$.
The potential energy at displacement $x$ is $U = \frac{1}{2} k x^2$, where $x = 3 \text{ cm} = 0.03 \text{ m}$.
The kinetic energy $K$ at displacement $x$ is given by $K = E - U = \frac{1}{2} k (A^2 - x^2)$.
Substituting the values:
$K = \frac{1}{2} \times 100 \times ((0.08)^2 - (0.03)^2)$
$K = 50 \times (0.0064 - 0.0009)$
$K = 50 \times 0.0055$
$K = 0.275 \text{ J}$

(C) The average energy $E$ of a one-dimensional harmonic oscillator in thermal equilibrium at temperature $T$ is given by the equipartition theorem as $E = k_B T$.
However,for a quantum harmonic oscillator,the average energy is given by $E = \frac{h\nu}{e^{h\nu/k_B T} - 1} + \frac{1}{2}h\nu$.
In the classical limit where $k_B T \gg h\nu$,the average energy is $E = k_B T$.
Assuming the question refers to the classical average energy per degree of freedom for a simple harmonic oscillator (which involves both kinetic and potential energy),the total average energy is $E = k_B T$.
Given $k_B = 1.38 \times 10^{-23} \ J K^{-1}$ and $T = 300 \ K$:
$E = (1.38 \times 10^{-23}) \times 300 = 4.14 \times 10^{-21} \ J$.

(B) The kinetic energy $(KE)$ of a simple harmonic oscillator is given by $KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$ and the total energy $(TE)$ is $TE = \frac{1}{2} m \omega^2 A^2$.
Given that $KE = 75\% \text{ of } TE$,we have:
$KE = \frac{3}{4} TE$
$\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{3}{4} (\frac{1}{2} m \omega^2 A^2)$
$A^2 - x^2 = \frac{3}{4} A^2$
$x^2 = A^2 - \frac{3}{4} A^2 = \frac{1}{4} A^2$
$x = \pm \frac{A}{2}$
For a particle starting from the mean position ($x=0$ at $t=0$),the displacement is given by $x = A \sin(\omega t)$.
Setting $x = \frac{A}{2}$,we get $\frac{A}{2} = A \sin(\omega t) \Rightarrow \sin(\omega t) = \frac{1}{2}$.
This implies $\omega t = \frac{\pi}{6}$.
Since $\omega = \frac{2\pi}{T}$,we have $\frac{2\pi}{T} t = \frac{\pi}{6}$.
$t = \frac{T}{12}$.
Given $T = 4 \ s$,$t = \frac{4}{12} = \frac{1}{3} \ s$.

(D) The potential energy $U$ of a body performing $SHM$ is given by $U = \frac{1}{2} k x^2$,where $x$ is the displacement from the mean position.
Given that the displacement $x = \frac{a}{2}$,where $a$ is the amplitude:
$U = \frac{1}{2} k \left(\frac{a}{2}\right)^2 = \frac{1}{2} k \frac{a^2}{4} = \frac{1}{8} k a^2$ ... $(i)$
The kinetic energy $K$ of the body is given by $K = \frac{1}{2} k (a^2 - x^2)$.
Substituting $x = \frac{a}{2}$:
$K = \frac{1}{2} k \left(a^2 - \frac{a^2}{4}\right) = \frac{1}{2} k \left(\frac{3 a^2}{4}\right) = \frac{3}{8} k a^2$ ... (ii)
Taking the ratio of kinetic energy to potential energy:
$\frac{K}{U} = \frac{\frac{3}{8} k a^2}{\frac{1}{8} k a^2} = 3$.

(A) The equation of motion for the particle is given by $y = 8 \cos [100 t + \pi / 4] \,cm$.
Comparing this with the standard $SHM$ equation $y = a \cos(\omega t + \phi)$, we get:
Amplitude $a = 8 \,cm = 8 \times 10^{-2} \,m$
Angular frequency $\omega = 100 \,rad/s$
Mass $m = 4 \,kg$
The maximum kinetic energy $(K_{max})$ in $SHM$ is given by the formula:
$K_{max} = \frac{1}{2} m \omega^2 a^2$
Substituting the values:
$K_{max} = \frac{1}{2} \times 4 \times (100)^2 \times (8 \times 10^{-2})^2$
$K_{max} = 2 \times 10000 \times 64 \times 10^{-4}$
$K_{max} = 2 \times 10000 \times 0.0064$
$K_{max} = 128 \,J$

(D) Let the amplitude of the simple harmonic motion be $a$. The displacement of the body from the mean position is $x = \frac{a}{N}$.
The kinetic energy $(KE)$ of the body is given by:
$KE = \frac{1}{2} m \omega^2 (a^2 - x^2) = \frac{1}{2} m \omega^2 \left(a^2 - \frac{a^2}{N^2}\right) = \frac{1}{2} m \omega^2 a^2 \left(1 - \frac{1}{N^2}\right) = \frac{1}{2} m \omega^2 a^2 \left(\frac{N^2 - 1}{N^2}\right) \quad (i)$
The potential energy $(PE)$ of the body is given by:
$PE = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 \left(\frac{a}{N}\right)^2 = \frac{1}{2} m \omega^2 \frac{a^2}{N^2} \quad (ii)$
Taking the ratio of $KE$ to $PE$:
$\frac{KE}{PE} = \frac{\frac{1}{2} m \omega^2 a^2 \left(\frac{N^2 - 1}{N^2}\right)}{\frac{1}{2} m \omega^2 \frac{a^2}{N^2}} = \frac{N^2 - 1}{1} = N^2 - 1$

(D) The energy of a particle executing $SHM$ is given by $E = \frac{1}{2} m \omega^2 A^2$, where $m$ is the mass, $\omega$ is the angular frequency, and $A$ is the amplitude.
For the first particle, $x_1 = 4 \sin (10t + \frac{\pi}{6})$, the amplitude $A_1 = 4$ and angular frequency $\omega_1 = 10$.
The energy is $E_1 = \frac{1}{2} m (10)^2 (4)^2 = \frac{1}{2} m (100)(16) = 800m$.
For the second particle, $x_2 = 5 \cos (\omega t)$, the amplitude $A_2 = 5$ and angular frequency is $\omega$.
The energy is $E_2 = \frac{1}{2} m \omega^2 (5)^2 = \frac{25}{2} m \omega^2$.
Given that the energies are equal, $E_1 = E_2$.
$800m = \frac{25}{2} m \omega^2$.
$1600 = 25 \omega^2$.
$\omega^2 = \frac{1600}{25} = 64$.
$\omega = 8 \text{ unit}$.

(D) The displacement of the particle executing $SHM$ is $y = 5 \sin \left(4t + \frac{\pi}{3}\right)$.
The velocity of the particle is given by $v = \frac{dy}{dt} = \frac{d}{dt} \left[5 \sin \left(4t + \frac{\pi}{3}\right)\right] = 5 \times 4 \cos \left(4t + \frac{\pi}{3}\right) = 20 \cos \left(4t + \frac{\pi}{3}\right)$.
Comparing the given equation with the standard $SHM$ equation $y = a \sin(\omega t + \phi)$, we get $\omega = 4 \text{ rad/s}$.
Since $\omega = \frac{2\pi}{T}$, we have $T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ s}$.
At $t = \frac{T}{4} = \frac{1}{4} \times \frac{\pi}{2} = \frac{\pi}{8} \text{ s}$, the velocity is:
$v = 20 \cos \left(4 \times \frac{\pi}{8} + \frac{\pi}{3}\right) = 20 \cos \left(\frac{\pi}{2} + \frac{\pi}{3}\right) = -20 \sin \left(\frac{\pi}{3}\right) = -20 \times \frac{\sqrt{3}}{2} = -10\sqrt{3} \text{ m/s}$.
The kinetic energy is $KE = \frac{1}{2}mv^2$.
Given $m = 2 \text{ g} = 2 \times 10^{-3} \text{ kg}$.
$KE = \frac{1}{2} \times (2 \times 10^{-3}) \times (-10\sqrt{3})^2 = 10^{-3} \times (100 \times 3) = 300 \times 10^{-3} = 0.3 \text{ J}$.

(B) Let the amplitude be $A = 6 \ cm$ and the displacement from the mean position be $x$.
The potential energy $(U)$ of a particle in simple harmonic motion is given by $U = \frac{1}{2} k x^2$.
The kinetic energy $(K)$ of the particle is given by $K = \frac{1}{2} k (A^2 - x^2)$.
Given the ratio of potential energy to kinetic energy is $U/K = 4/5$.
Substituting the expressions,we get $\frac{\frac{1}{2} k x^2}{\frac{1}{2} k (A^2 - x^2)} = \frac{4}{5}$.
This simplifies to $\frac{x^2}{A^2 - x^2} = \frac{4}{5}$.
Cross-multiplying gives $5x^2 = 4(A^2 - x^2) = 4A^2 - 4x^2$.
Rearranging terms,$9x^2 = 4A^2$,which implies $x^2 = \frac{4}{9} A^2$.
Taking the square root,$x = \frac{2}{3} A$.
Given $A = 6 \ cm$,we find $x = \frac{2}{3} \times 6 \ cm = 4 \ cm$.

(A) The kinetic energy $K$ of a particle in simple harmonic motion at a displacement $x$ is given by $K = \frac{1}{2} k(A^2 - x^2)$,where $A$ is the amplitude.
Maximum kinetic energy $K_{max} = \frac{1}{2} kA^2$.
Given that at $x = 4 \,cm$,$K = \frac{1}{3} K_{max}$.
Substituting the expressions: $\frac{1}{2} k(A^2 - 4^2) = \frac{1}{3} (\frac{1}{2} kA^2)$.
Dividing both sides by $\frac{1}{2} k$: $A^2 - 16 = \frac{A^2}{3}$.
Rearranging the terms: $A^2 - \frac{A^2}{3} = 16$.
$\frac{2A^2}{3} = 16$.
$A^2 = \frac{16 \times 3}{2} = 24$.
$A = \sqrt{24} = 2\sqrt{6} \,cm$.

(D) The equation for position in $S.H.M$ is given by $x(t) = A \cos(\omega t + \phi)$,where $A = 5$ is the amplitude.
The total mechanical energy $E$ of an $S.H.M$ system is given by $E = \frac{1}{2} k A^2 = 100 \ J$.
The potential energy $U$ at any time $t$ is given by $U = \frac{1}{2} k x^2$.
At $t = 0$,the position is $x(0) = 5 \cos\left(0 + \frac{\pi}{4}\right) = 5 \cos\left(\frac{\pi}{4}\right) = 5 \times \frac{1}{\sqrt{2}}$.
Substituting $x(0)$ into the potential energy formula:
$U(0) = \frac{1}{2} k \left(5 \times \frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} k \left(\frac{25}{2}\right) = \frac{1}{4} k (25)$.
Since $E = \frac{1}{2} k A^2 = \frac{1}{2} k (5)^2 = \frac{25}{2} k = 100 \ J$,we find $k = \frac{200}{25} = 8 \ J/m^2$.
Substituting $k = 8$ into the expression for $U(0)$:
$U(0) = \frac{1}{4} \times 8 \times 25 = 2 \times 25 = 50 \ J$.

(C) The given equation for position in $SHM$ is $x(t) = 2 \cos \left(\frac{\pi}{15} t - \frac{\pi}{2}\right)$.
Comparing this with the standard equation $x(t) = A \cos(\omega t + \phi)$,we get the angular frequency $\omega = \frac{\pi}{15} \text{ rad/s}$.
The time period of the $SHM$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/15} = 30 \text{ s}$.
The kinetic energy of a particle in $SHM$ is given by $KE = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t + \phi)$.
Using the identity $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$,the kinetic energy varies with a frequency of $2\omega$.
Therefore,the time period of the kinetic energy is $T_{KE} = \frac{T}{2} = \frac{30}{2} = 15 \text{ s}$.

(C) The total energy $E$ of a particle in simple harmonic motion is given by $E = \frac{1}{2} m \omega^2 A^2$.
Potential energy $U$ at displacement $x$ is given by $U = \frac{1}{2} m \omega^2 x^2$.
According to the problem,the potential energy is half of the total energy,so $U = \frac{E}{2}$.
Substituting the expressions,we get $\frac{1}{2} m \omega^2 x^2 = \frac{1}{2} \left( \frac{1}{2} m \omega^2 A^2 \right)$.
Simplifying the equation,we get $x^2 = \frac{A^2}{2}$.
Taking the square root on both sides,we find $x = \pm \frac{A}{\sqrt{2}}$.

$(B)$ The displacement of a particle executing $SHM$ is given by $y = a \sin(\omega t)$.
The velocity of the particle is $u = \frac{dy}{dt} = a\omega \cos(\omega t)$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mu^2 = \frac{1}{2}m(a\omega \cos(\omega t))^2 = \frac{1}{2}m\omega^2 a^2 \cos^2(\omega t)$.
Using the trigonometric identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$, we get $K = \frac{1}{4}m\omega^2 a^2 (1 + \cos(2\omega t))$.
Since the frequency of the $SHM$ is $v = \frac{\omega}{2\pi}$, the frequency of the kinetic energy oscillation is determined by the term $\cos(2\omega t)$, which is $v' = \frac{2\omega}{2\pi} = 2v$.
Therefore, the kinetic energy changes periodically with a frequency of $2v$.

(B) The potential energy of a simple harmonic oscillator is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(\omega t)$.
Potential energy is maximum when $\sin^2(\omega t) = 1$,which occurs at the extreme positions.
For a particle starting from the mean position ($x=0$ at $t=0$),it reaches the first extreme position $(x=A)$ at time $t = T/4$.
Given that the maximum potential energy occurs at $t = T / (2 \beta)$,we equate the two expressions:
$T / (2 \beta) = T / 4$.
By comparing the denominators,we get $2 \beta = 4$,which implies $\beta = 2$.

(D) The kinetic energy $(K.E)$ of a simple pendulum is given by $K.E = \frac{1}{2} m v^2$.
The velocity $v$ of a simple pendulum performing simple harmonic motion $(SHM)$ is $v = A\omega \cos(\omega t + \phi)$.
Thus,$K.E = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t + \phi)$.
Since $K.E \propto \cos^2(\omega t)$,the graph is a periodic function with a frequency double that of the pendulum's motion,meaning it completes two cycles in the time the pendulum completes one cycle $(T)$.
At the mean position $(t=0)$,the velocity is maximum,so kinetic energy is maximum. The graph that shows maximum kinetic energy at $t=0$ and completes two cycles in time $T$ is Graph $D$.