Simple Harmonic Motion

The potential energy $(U)$ of a particle executing simple harmonic motion $(S H M)$ is given by :

$ \mathrm{U}=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2 $

Substituting the displacement equation $\mathrm{x}(\mathrm{t})=\mathrm{A} \sin \omega \mathrm{t}$ :

$ \mathrm{U}(\mathrm{t})=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2 \sin ^2 \omega \mathrm{t} $

The potential energy $U$ is maximum when the displacement $x$ is maximum $(x= \pm A)$.

This occurs when :

$ \sin ^2 \omega t=1 \Rightarrow \sin \omega t= \pm 1 $

This occurs for a particle starting from the mean position ( $t=0$ ) is at :

$ \omega \mathrm{t}=\frac{\pi}{2} $

Using the relationship between angular frequency $\omega$ and time period (T), $\omega=\frac{2 \pi}{\mathrm{~T}}$

$ \Rightarrow\left(\frac{2 \pi}{T}\right) t=\frac{\pi}{2} \Rightarrow t=\frac{T}{4} $

It is given that $\mathrm{t}=\frac{\mathrm{T}}{2 \beta}$,

$ \frac{T}{4}=\frac{T}{2 \beta} \Rightarrow \beta=2 $

Therefore, the value of $\beta$ is 2 . So, the correct answer is 2 .

To find the value of $x$ that represents the time period of oscillation in this simple harmonic motion (SHM) scenario, we first recall the general formula for the time period ($T$) of a mass-spring system undergoing SHM, which is given by:

$T = 2\pi \sqrt{\frac{m}{k}}$

Here,

$m$ is the mass of the particle, which is $0.50 \, \mathrm{kg}$ in this case,

$k$ is the force constant of the spring or the spring constant, which is given as $50 \, \mathrm{Nm^{-1}}$,

and $T$ represents the time period of oscillation.

Given in the problem, $T = \frac{x}{35} \, \mathrm{s}$ and we are provided with the approximation $\pi = \frac{22}{7}$.

Substituting the given values into the formula for $T$:

$\frac{x}{35} = 2 \times \frac{22}{7} \times \sqrt{\frac{0.50}{50}}$

To simplify this, we first calculate the square root:

$\sqrt{\frac{0.50}{50}} = \sqrt{\frac{1}{100}} = \frac{1}{10}$

Substituting back, we get:

$\frac{x}{35} = 2 \times \frac{22}{7} \times \frac{1}{10}$

Multiplying the terms on the right side:

$\frac{x}{35} = \frac{44}{70}$

$\frac{x}{35} = \frac{22}{35}$

Multiplying both sides by $35$ to solve for $x$:

$x = 22$

Therefore, the value of $x$ that represents the time period of oscillation is $22$ seconds.

Let's begin by understanding the equations related to simple harmonic motion (SHM). For a particle executing SHM, the position $x$, velocity $v$, and acceleration $a$ are given by the following equations:

1. Position: $x = A \cos(\omega t + \phi)$

2. Velocity: $v = -A \omega \sin(\omega t + \phi)$

3. Acceleration: $a = -A \omega^2 \cos(\omega t + \phi)$

Here, $A$ is the amplitude of the motion, $\omega$ is the angular frequency, and $\phi$ is the phase constant.

Given the magnitudes at a certain instant:

$x = 4 \, \mathrm{m}$

$v = 2 \, \mathrm{ms}^{-1}$

$a = 16 \, \mathrm{ms}^{-2}$

Using the acceleration equation:

$a = -A \omega^2 \cos(\omega t + \phi)$

Since we’re given the magnitude of the acceleration, we remove the negative sign:

$16 = A \omega^2 \cos(\omega t + \phi)$

Using the position equation:

$x = A \cos(\omega t + \phi)$

We already know $x = 4 \, \mathrm{m}$, so:

$4 = A \cos(\omega t + \phi)$

From these two equations, we know:

$A \omega^2 \cos(\omega t + \phi) = 16$

$A \cos(\omega t + \phi) = 4$

Therefore:

$A \omega^2 \cdot 4/A = 16$

$4 \omega^2 = 16$

$\omega^2 = 4$

$\omega = 2 \, \mathrm{rad/s}$

Next, using the velocity equation:

$v = -A \omega \sin(\omega t + \phi)$

Again, we consider the magnitude:

$2 = A \cdot 2 \sin(\omega t + \phi)$

$2 = 2A \sin(\omega t + \phi)$

$\sin(\omega t + \phi) = \dfrac{1}{A}$

We know from the position equation that:

$\cos(\omega t + \phi) = \dfrac{4}{A}$

Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, we get:

$\left(\dfrac{1}{A}\right)^2 + \left(\dfrac{4}{A}\right)^2 = 1$

$\dfrac{1}{A^2} + \dfrac{16}{A^2} = 1$

$\dfrac{17}{A^2} = 1$

$A^2 = 17$

$A = \sqrt{17} \, \mathrm{m}$

Therefore, the amplitude of the motion is $\sqrt{17} \, \mathrm{m}$, meaning $x$ is 17.

To solve for the amplitude of oscillation, we start by using the properties of simple harmonic motion (SHM). In SHM, the total energy of the system is conserved and is given by the sum of kinetic energy (KE) and potential energy (PE).

Given:

• Mass $m = 0.2 \ \mathrm{kg}$
• Frequency $f = \left(\frac{25}{\pi}\right) \ \mathrm{Hz}$
• Position $x = 0.04 \ \mathrm{m}$
• KE at $x = 0.04 \ \mathrm{m}$ is $0.5 \ \mathrm{J}$
• PE at $x = 0.04 \ \mathrm{m}$ is $0.4 \ \mathrm{J}$

The total mechanical energy (E) of the SHM system can be found by summing the given kinetic and potential energies:

$ E = KE + PE = 0.5 \ \mathrm{J} + 0.4 \ \mathrm{J} = 0.9 \ \mathrm{J} $

For simple harmonic motion, the total energy (E) is also related to the amplitude (A) by the following formula:

$ E = \frac{1}{2} k A^2 $

where $k$ is the spring constant. First, we need to find the angular frequency $\omega$:

$ \omega = 2 \pi f = 2 \pi \left(\frac{25}{\pi}\right) \ \mathrm{Hz} = 50 \ \mathrm{rad/s} $

The spring constant $k$ can be calculated using the relationship between $m$, $\omega$, and $k$:

$ \omega = \sqrt{\frac{k}{m}} \Rightarrow k = m \omega^2 = 0.2 \times (50)^2 = 500 \ \mathrm{N/m} $

Now, substituting $k$ back into the energy equation, we solve for the amplitude $A$:

$ 0.9 = \frac{1}{2} \times 500 \times A^2 \Rightarrow A^2 = \frac{0.9 \times 2}{500} \Rightarrow A^2 = \frac{1.8}{500} \Rightarrow A^2 = 0.0036 \Rightarrow A = \sqrt{0.0036} = 0.06 \ \mathrm{m} $

Converting $A$ from meters to centimeters:

$ A = 0.06 \ \mathrm{m} \times 100 = 6 \ \mathrm{cm} $

Thus, the amplitude of oscillation is $6 \ \mathrm{cm}$.

For a particle performing simple harmonic motion (SHM), the maximum velocity $v_{max}$ can be calculated using the formula:

$v_{max} = A\omega$

where $A$ is the amplitude of the motion and $\omega$ is the angular frequency. The angular frequency $\omega$ is related to the time period $T$ by the formula:

$\omega = \frac{2\pi}{T}$

Given:

• Amplitude, $A = 0.06 \, \mathrm{m}$


• Time period, $T = 3.14 \, \mathrm{s}$

First, we find the angular frequency:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{3.14}$

Substituting $\omega$ and $A$ in the formula for $v_{max}$:

$v_{max} = A\omega = 0.06 \times \frac{2\pi}{3.14}$

$v_{max} = 0.06 \times \frac{2 \times 3.14}{3.14}$

$v_{max} = 0.06 \times 2$

$v_{max} = 0.12 \, \mathrm{m/s}$

To convert meters per second to centimeters per second, we use the conversion factor $1 \, \mathrm{m/s} = 100 \, \mathrm{cm/s}$. Therefore,

$v_{max} = 0.12 \, \mathrm{m/s} \times 100 \, \mathrm{cm/m} = 12 \, \mathrm{cm/s}$

Thus, the maximum velocity of the particle is $12 \, \mathrm{cm/s}$.

The displacement of a particle executing Simple Harmonic Motion (SHM) can be expressed as:

$x = A \sin(\omega t + \phi)$

Where:

$A$ is the amplitude of the SHM,

$\omega$ is the angular frequency,

$t$ is the time,

$\phi$ is the phase constant (phase angle at $t = 0$).

In the given equation, $x = 10 \sin(\omega t + \frac{\pi}{3})$ m, the amplitude $A = 10$ m and the phase constant $\phi = \frac{\pi}{3}$. The time period $T = 3.14$ s is given, from which we can find the angular frequency $\omega$ using the relationship:

$\omega = \frac{2\pi}{T}$

Substituting the given $T = 3.14$ s:

$\omega = \frac{2\pi}{3.14} \approx 2 \, \text{rad/s}$

To find the velocity of the particle, we differentiate the displacement $x$ with respect to time $t$. The derivative of the displacement gives the velocity:

$v = \frac{dx}{dt}$

So, for $x = 10 \sin(\omega t + \frac{\pi}{3})$:

$v = \frac{d}{dt}[10 \sin(\omega t + \frac{\pi}{3})]$

Applying differentiation, we get:

$v = 10\omega \cos(\omega t + \frac{\pi}{3})$

Plug in the value of $\omega = 2$ rad/s and evaluate it at $t = 0$ to find the initial velocity:

$v = 10 \cdot 2 \cos(2 \cdot 0 + \frac{\pi}{3})$

$v = 20 \cos(\frac{\pi}{3})$

$\cos(\frac{\pi}{3}) = \frac{1}{2}$, therefore:

$v = 20 \cdot \frac{1}{2} = 10 \, \text{m/s}$

Thus, the velocity of the particle at $t = 0$ is $10$ m/s.

Let's start by considering the formula for the frequency of a mass on a spring (a simple harmonic oscillator):

$ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $

Where:

• $ f $ is the frequency of oscillation
• $ k $ is the spring constant
• $ m $ is the mass suspended from the spring

When the mass $ m $ is suspended, the frequency $ f_1 $ is:

$ f_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $

When the mass $ 9m $ is suspended, the frequency $ f_2 $ is:

$ f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{9m}} $

We can simplify the square root by taking the 9 inside the root as $3^2$, which gives:

$ f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{(3^2)m}} $

$ f_2 = \frac{1}{2\pi} \frac{1}{3} \sqrt{\frac{k}{m}} $

The ratio of $ \frac{f_1}{f_2} $ is therefore:

$ \frac{f_1}{f_2} = \frac{\frac{1}{2\pi} \sqrt{\frac{k}{m}}}{\frac{1}{2\pi} \frac{1}{3} \sqrt{\frac{k}{m}}} $

$ \frac{f_1}{f_2} = \frac{1}{\frac{1}{3}} $

$ \frac{f_1}{f_2} = 3 $

So the value of $ \frac{f_1}{f_2} $ is $3$.

$\mathrm{k}_{\mathrm{eq}}=\frac{2 \mathrm{k} \cdot \mathrm{k}}{3 \mathrm{k}}+\mathrm{k}=\frac{5 \mathrm{k}}{3}$

Angular frequency of oscillation $(\omega)=\sqrt{\frac{\mathrm{k}_{\mathrm{eq}}}{\mathrm{m}}}$

$(\omega)=\sqrt{\frac{5 \mathrm{k}}{3 \mathrm{~m}}}$

Period of oscillation $(\tau)=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{3 \mathrm{~m}}{5 \mathrm{k}}}$

$=\pi \sqrt{\frac{12 \mathrm{~m}}{5 \mathrm{k}}}$

To find the new amplitude of the motion when the speed is increased to three times at a given displacement, we use the concepts of simple harmonic motion (SHM) and its formulas.

In SHM, the velocity $v$ of a particle at a displacement $x$ from the mean position can be given by the formula:

$v = \omega \sqrt{A^2 - x^2}$

where:

• $\omega$ is the angular frequency of the motion,
• $A$ is the amplitude, and
• $x$ is the displacement at that instance.

Given:

• Displacement at the instance, $x = \frac{2A}{3}$,
• Initial velocity is increased to three times at this displacement.

Thus, let's find the initial velocity $v$ at $x = \frac{2A}{3}$:

$v = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\sqrt{5}A\omega}{3}$

With the velocity increased to three times, the new velocity $v'$ becomes:

$v' = 3v = 3 \times \frac{\sqrt{5}A\omega}{3} = \sqrt{5}A\omega$

For the new amplitude $A'$, the velocity $v'$ at the same displacement $x$ is:

$v' = \omega \sqrt{{A'}^2 - \left(\frac{2A}{3}\right)^2}$

Setting the expressions for $v'$ equal gives:

$\sqrt{5}A\omega = \omega \sqrt{{A'}^2 - \frac{4A^2}{9}}$

$\sqrt{5}A = \sqrt{{A'}^2 - \frac{4A^2}{9}}$

Solving for $A'$ gives:

${A'}^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9} = \frac{49A^2}{9}$

$A' = \sqrt{\frac{49A^2}{9}} = \frac{7A}{3}$

Therefore, the new amplitude of the motion is $\frac{7A}{3}$, which means the value of $n$ is 7.

From phasor diagram particle has to move from $\mathrm{P}$ to $\mathrm{Q}$ in a circle of radius equal to amplitude of SHM.

$\begin{aligned} & \cos \phi=\frac{\frac{\sqrt{3} \mathrm{~A}}{2}}{\mathrm{~A}}=\frac{\sqrt{3}}{2} \\ & \phi=\frac{\pi}{6} \end{aligned}$

Now, $\frac{\pi}{6}=\omega \mathrm{t}$

$\begin{aligned} & \frac{\pi}{6}=\frac{2 \pi}{T} t \\ & \frac{\pi}{6}=\frac{2 \pi}{6 \pi} t \end{aligned}$

$\mathrm{t}=\frac{\pi}{2}$

So, $x=2$

$\begin{aligned} & \text { Let total energy }=\mathrm{E}=\frac{1}{2} \mathrm{KA}^2 \\ & \mathrm{U}=\frac{1}{2} \mathrm{~K}\left(\frac{\mathrm{A}}{3}\right)^2=\frac{\mathrm{KA}^2}{2 \times 9}=\frac{\mathrm{E}}{9} \\ & \mathrm{KE}=\mathrm{E}-\frac{\mathrm{E}}{9}=\frac{8 \mathrm{E}}{9} \\ & \text { Ratio } \frac{\text { Total }}{\mathrm{KE}}=\frac{\mathrm{E}}{\frac{8 \mathrm{E}}{9}}=\frac{9}{8} \\ & \mathrm{x}=9 \end{aligned}$

$\begin{aligned} & \mathrm{V}_{\text {at mean position }}=\mathrm{A} \omega \Rightarrow 10=4 \omega \\ & \quad \omega=\frac{5}{2} \\ & \mathrm{~V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \\ & 5=\frac{5}{2} \sqrt{4^2-\mathrm{x}^2} \Rightarrow \mathrm{x}^2=16-4 \\ & \mathrm{x}=\sqrt{12} \mathrm{~cm} \end{aligned}$

To find the maximum force exerted by the spring on the block, we can use Hooke's law and the properties of simple harmonic motion.

First, let's find the angular frequency $\omega$:

$\omega = \frac{2\pi}{T}$

where $T = 3.14\,\mathrm{s}$ is the time period.

$\omega = \frac{2\pi}{3.14} \approx 2\,\mathrm{rad/s}$

Now, let's find the maximum velocity $v_{max}$ of the block:

$v_{max} = \omega A$

where $A = 1\,\mathrm{m}$ is the amplitude.

$v_{max} = 2\,\mathrm{rad/s} \times 1\,\mathrm{m} = 2\,\mathrm{m/s}$

Next, we can find the spring constant $k$ using the mass of the block $m = 5\,\mathrm{kg}$ and the angular frequency $\omega$:

$\omega^2 = \frac{k}{m} \Rightarrow k = m\omega^2$

$k = 5\,\mathrm{kg} \times (2\,\mathrm{rad/s})^2 = 20\,\mathrm{N/m}$

Finally, we can find the maximum force exerted by the spring on the block. At maximum displacement, the force is given by Hooke's law:

$F_{max} = kA$

$F_{max} = 20\,\mathrm{N/m} \times 1\,\mathrm{m} = 20\,\mathrm{N}$

The maximum force exerted by the spring on the block is $20\,\mathrm{N}$.

Given the amplitude $A$ and the length $l$ of the pendulum, we can find the maximum angular displacement $\theta_0$:

$ \sin \theta_0 = \frac{A}{l} = \frac{10}{100} = \frac{1}{10} $

By conservation of energy, the following equation holds:

$ \frac{1}{2} m v^2 = m g l(1 - \cos \theta) $

The maximum tension occurs at the mean position (i.e., when the pendulum is vertical). At this point, we have:

$ \begin{aligned} & T - mg = \frac{m v^2}{l} \ & \Rightarrow T = mg + \frac{m v^2}{l} \end{aligned} $

Substituting the conservation of energy equation, we get:

$ \begin{aligned} & T = mg + 2 m g(1 - \cos \theta) \\\\ & = mg\left[1 + 2\left(1 - \sqrt{1 - \sin^2 \theta}\right)\right] \\\\ & = mg\left[3 - 2 \sqrt{1 - \frac{1}{100}}\right] \\\\ & = \frac{250}{1000} \times 10\left[3 - 2\left(1 - \frac{1}{200}\right)\right] = \frac{101}{40} \\\\ & \therefore x = 101 \end{aligned} $

So, the value of $x$ is 101.