The displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t)=A \sin \omega t$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t=T / 2 \beta$. The value of $\beta$ is $\_\_\_\_$ .
Explanation:
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 .
A particle of mass $0.50 \mathrm{~kg}$ executes simple harmonic motion under force $F=-50(\mathrm{Nm}^{-1}) x$. The time period of oscillation is $\frac{x}{35} s$. The value of $x$ is _________.
(Given $\pi=\frac{22}{7}$)
Explanation:
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.
The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 \mathrm{~m}, 2 \mathrm{~ms}^{-1}$ and $16 \mathrm{~ms}^{-2}$ at a certain instant. The amplitude of the motion is $\sqrt{x}, \mathrm{~m}$ where $x$ is _________.
Explanation:
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.
An object of mass $0.2 \mathrm{~kg}$ executes simple harmonic motion along $x$ axis with frequency of $\left(\frac{25}{\pi}\right) \mathrm{Hz}$. At the position $x=0.04 \mathrm{~m}$ the object has kinetic energy $0.5 \mathrm{~J}$ and potential energy $0.4 \mathrm{~J}$. The amplitude of oscillation is ________ $\mathrm{cm}$.
Explanation:
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 $6 \ \mathrm{cm}$.
A particle is doing simple harmonic motion of amplitude $0.06 \mathrm{~m}$ and time period $3.14 \mathrm{~s}$. The maximum velocity of the particle is _________ $\mathrm{cm} / \mathrm{s}$.
Explanation:
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 SHM is given by $x=10 \sin \left(w t+\frac{\pi}{3}\right) m$. The time period of motion is $3.14 \mathrm{~s}$. The velocity of the particle at $t=0$ is _______ $\mathrm{m} / \mathrm{s}$.
Explanation:
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.
Explanation:
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 $3$.
The time period of simple harmonic motion of mass $M$ in the given figure is $\pi \sqrt{\frac{\alpha M}{5 k}}$, where the value of $\alpha$ is _________.
Explanation:
$\mathrm{k}_{\mathrm{eq}}=\frac{2 \mathrm{k} \cdot \mathrm{k}}{3 \mathrm{k}}+\mathrm{k}=\frac{5 \mathrm{k}}{3}$
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 $(\tau)=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{3 \mathrm{~m}}{5 \mathrm{k}}}$
$=\pi \sqrt{\frac{12 \mathrm{~m}}{5 \mathrm{k}}}$
A particle performs simple harmonic motion with amplitude $A$. Its speed is increased to three times at an instant when its displacement is $\frac{2 A}{3}$. The new amplitude of motion is $\frac{n A}{3}$. The value of $n$ is ___________.
Explanation:
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.
A simple harmonic oscillator has an amplitude $A$ and time period $6 \pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x=$ A to $x=\frac{\sqrt{3}}{2}$ A will be $\frac{\pi}{x} \mathrm{~s}$, where $x=$ _________.
Explanation:
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, $x=2$
When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is $\frac{x}{8}$, where $x=$ _________.
Explanation:
$\begin{aligned} & \text { Let total energy }=\mathrm{E}=\frac{1}{2} \mathrm{KA}^2 \\ & \mathrm{U}=\frac{1}{2} \mathrm{~K}\left(\frac{\mathrm{A}}{3}\right)^2=\frac{\mathrm{KA}^2}{2 \times 9}=\frac{\mathrm{E}}{9} \\ & \mathrm{KE}=\mathrm{E}-\frac{\mathrm{E}}{9}=\frac{8 \mathrm{E}}{9} \\ & \text { Ratio } \frac{\text { Total }}{\mathrm{KE}}=\frac{\mathrm{E}}{\frac{8 \mathrm{E}}{9}}=\frac{9}{8} \\ & \mathrm{x}=9 \end{aligned}$
A particle executes simple harmonic motion with an amplitude of $4 \mathrm{~cm}$. At the mean position, velocity of the particle is $10 \mathrm{~cm} / \mathrm{s}$. The distance of the particle from the mean position when its speed becomes $5 \mathrm{~cm} / \mathrm{s}$ is $\sqrt{\alpha} \mathrm{~cm}$, where $\alpha=$ ________.
Explanation:
$\begin{aligned} & \mathrm{V}_{\text {at mean position }}=\mathrm{A} \omega \Rightarrow 10=4 \omega \\ & \quad \omega=\frac{5}{2} \\ & \mathrm{~V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \\ & 5=\frac{5}{2} \sqrt{4^2-\mathrm{x}^2} \Rightarrow \mathrm{x}^2=16-4 \\ & \mathrm{x}=\sqrt{12} \mathrm{~cm} \end{aligned}$
At a given point of time the value of displacement of a simple harmonic oscillator is given as $\mathrm{y}=\mathrm{A} \cos \left(30^{\circ}\right)$. If amplitude is $40 \mathrm{~cm}$ and kinetic energy at that time is $200 \mathrm{~J}$, the value of force constant is $1.0 \times 10^{x} ~\mathrm{Nm}^{-1}$. The value of $x$ is ____________.
Explanation:
$x = A \sin(\omega t + \phi)$
At the given time, we have:
$\omega t + \phi = 30^\circ$
Given the amplitude $A = 40 \,\text{cm}$ and the displacement $x = 40 \times \frac{\sqrt{3}}{2} \,\text{cm} = 20\sqrt{3} \,\text{cm}$, we can write the kinetic energy, $KE$, as:
$KE = \frac{1}{2}k(A^2 - x^2) = 200$
Now, substitute the values for $A$ and $x$:
$200 = \frac{1}{2}k\left(\frac{1600 - 1200}{100 \times 100}\right)$
Simplify the equation:
$400 \times 100 \times 100 = k \times 400$
Solve for the force constant, $k$:
$k = 10^4 \,\text{Nm}^{-1}$
Given that the force constant is expressed as $k = 1.0 \times 10^x \,\text{Nm}^{-1}$, comparing the values, we get:
$1.0 \times 10^x = 10^4$
Thus, the value of $x$ is $4$.
A rectangular block of mass $5 \mathrm{~kg}$ attached to a horizontal spiral spring executes simple harmonic motion of amplitude $1 \mathrm{~m}$ and time period $3.14 \mathrm{~s}$. The maximum force exerted by spring on block is _________ N
Explanation:
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 $20\,\mathrm{N}$.
A simple pendulum with length $100 \mathrm{~cm}$ and bob of mass $250 \mathrm{~g}$ is executing S.H.M. of amplitude $10 \mathrm{~cm}$. The maximum tension in the string is found to be $\frac{x}{40} \mathrm{~N}$. The value of $x$ is ___________.
Explanation:
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.
The amplitude of a particle executing SHM is $3 \mathrm{~cm}$. The displacement at which its kinetic energy will be $25 \%$ more than the potential energy is: __________ $\mathrm{cm}$
Explanation:
$ \begin{aligned} & K=1.25 U \\\\ & \Rightarrow K+\frac{K}{1.25}=K_{\max } \\\\ & \Rightarrow \frac{9}{5} K=K_{\max } \\\\ & \Rightarrow \frac{9}{5} \frac{1}{2} m v^{2}=\frac{1}{2} m v_{\max }^{2} \\\\ & \Rightarrow \frac{9}{5}\left[\omega \sqrt{A^{2}-x^{2}}\right]^{2}=\omega^{2} A^{2} \\\\ & \Rightarrow 9\left(A^{2}-x^{2}\right)=5 A^{2} \\\\ & \Rightarrow x^{2}=\frac{4 A^{2}}{9} \\\\ & \Rightarrow x=\frac{2 A}{3} \\\\ & \Rightarrow x=2 \mathrm{~cm} \end{aligned} $
In the figure given below, a block of mass $M=490 \mathrm{~g}$ placed on a frictionless table is connected with two springs having same spring constant $\left(\mathrm{K}=2 \mathrm{~N} \mathrm{~m}^{-1}\right)$. If the block is horizontally displaced through '$\mathrm{X}$' $\mathrm{m}$ then the number of complete oscillations it will make in $14 \pi$ seconds will be _____________.
Explanation:
$=2 \times 2=4 \mathrm{~N} / \mathrm{m}$
$ \mathrm{m}=490 \mathrm{gm} $
$ \begin{aligned} & =0.49 \mathrm{~kg} \end{aligned} $
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{Keff}}}=2 \pi \sqrt{\frac{0.49 \mathrm{~kg}}{4}}$
$ =2 \pi \sqrt{\frac{49}{400}}=2 \pi \frac{7}{20}=\frac{7 \pi}{10} $
No. of oscillation in the $14 \pi$ is
$ \mathrm{N}=\frac{\text { time }}{\mathrm{T}}=\frac{14 \pi}{7 \pi / 10}=20 $
Explanation:
$4{v^2} = 50 - {x^2}$
or $v = {1 \over 2}\sqrt {50 - {x^2}} $
Comparing the above equation with $v = \omega \sqrt {{A^2} - {x^2}} $
$ \Rightarrow \omega = {1 \over 2}$
& $A = \sqrt {50} $
so ${{2\pi } \over T} = {1 \over 2}$
$ \Rightarrow T = 4\pi \sec $
$ = 4 \times {{22} \over 7}\sec $
$T = {{88} \over 7}\sec $
so $x = 88$
The general displacement of a simple harmonic oscillator is $x = A\sin \omega t$. Let T be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when $t = {T \over \beta }$. The value of $\beta$ is ______________.
Explanation:
$U = {1 \over 2}m{\omega ^2}{A^2}{\sin ^2}\omega t$
So, ${{dU} \over {dt}} = {{m{\omega ^3}{A^2}} \over 2}\sin 2\omega t$
This value will be maximum when $\sin 2\omega t = 1$
or $2\omega t = {\pi \over 2}$
$2 \times {{2\pi } \over T}t = {\pi \over 2}$
$ \Rightarrow t = {T \over 8}$
So $\beta = 8$
A particle of mass 250 g executes a simple harmonic motion under a periodic force $\mathrm{F}=(-25~x)\mathrm{N}$. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ___________ cm.
Explanation:
$F = - 25x$
$.250{{{d^2}x} \over {d{t^2}}} = - 25x$
${{{d^2}x} \over {d{t^2}}} = - 100x$
$ \Rightarrow \omega = 10$ rad/sec
& $\omega A = {v_{\max }}$
$10\,A = 4$
$ \Rightarrow A = 0.4$ m
$ = 40$ cm
A mass m attached to free end of a spring executes SHM with a period of 1s. If the mass is increased by 3 kg the period of the oscillation increases by one second, the value of mass m is ___________ kg.
Explanation:
Finally
$ 2 \pi \sqrt{\frac{m+3}{k}}=1+1=2 $
Equation $\frac{(1)}{(2)}$ gives
$\sqrt{\frac{m}{m+3}}=\frac{1}{2}$
$\therefore m=1 \mathrm{~kg}$
A block of a mass 2 kg is attached with two identical springs of spring constant 20 N/m each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is $\frac{\pi}{\sqrt x}$ in SI unit. The value of $x$ is ____________.
Explanation:
So $T=2 \pi \sqrt{\frac{m}{K_{n e t}}}$
$ \begin{aligned} & =2 \pi \sqrt{\frac{2}{40}} \\\\ & =\frac{\pi}{\sqrt{5}} \end{aligned} $
$\therefore x=5$
The metallic bob of simple pendulum has the relative density 5. The time period of this pendulum is $10 \mathrm{~s}$. If the metallic bob is immersed in water, then the new time period becomes $5 \sqrt{x}$ s. The value of $x$ will be ________.
Explanation:
$\mathrm{mg}^{\prime}=\mathrm{mg}-\mathrm{F}_{\mathrm{B}}$
$\mathrm{g}^{\prime}=\frac{\mathrm{mg}-\mathrm{F}_{\mathrm{B}}}{\mathrm{F}_{\mathrm{B}}}$
$=\frac{\rho_{\mathrm{B}} \mathrm{Vg}-\rho_{\mathrm{w}} \mathrm{Vg}}{\rho_{\mathrm{B}} \mathrm{V}}$
$=\left(\frac{\rho_{\mathrm{B}}-\rho_{\mathrm{w}}}{\rho_{\mathrm{B}}}\right) \mathrm{g}$
$=\frac{5-1}{5} \times \mathrm{g}$
$=\frac{4}{5} \mathrm{~g}$
We know, $T =2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$
$\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{\mathrm{g}^{\prime}}}=\sqrt{\frac{\mathrm{g}}{5} \mathrm{~g}}=\sqrt{\frac{5}{4}}$
$\mathrm{~T}^{\prime}=\mathrm{T} \sqrt{\frac{5}{4}}=\frac{10}{2} \sqrt{5}$
$\mathrm{~T}^{\prime}=5 \sqrt{5}$
The potential energy of a particle of mass $4 \mathrm{~kg}$ in motion along the x-axis is given by $\mathrm{U}=4(1-\cos 4 x)$ J. The time period of the particle for small oscillation $(\sin \theta \simeq \theta)$ is $\left(\frac{\pi}{K}\right) s$. The value of $\mathrm{K}$ is _________.
Explanation:
$U = 4(1 - \cos 4x)$
$ \Rightarrow F = - {{dU} \over {dx}} = - (4)(4\sin 4x)$
$ = - 16\sin 4x$
as small x
$F = - 16(4x) = - 64x \equiv - kx$
$T = 2\pi \sqrt {{m \over k}} = 2\pi \sqrt {{4 \over {64}}} = {\pi \over 2}$
$ \Rightarrow K = 2$
A mass $0.9 \mathrm{~kg}$, attached to a horizontal spring, executes SHM with an amplitude $\mathrm{A}_{1}$. When this mass passes through its mean position, then a smaller mass of $124 \mathrm{~g}$ is placed over it and both masses move together with amplitude $A_{2}$. If the ratio $\frac{A_{1}}{A_{2}}$ is $\frac{\alpha}{\alpha-1}$, then the value of $\alpha$ will be ___________.
Explanation:
$(0.9){A_1}\sqrt {{K \over {0.9}}} = (0.9 + 0.124){A_2}\sqrt {{K \over {0.9 + 0.124}}} $
${{{A_1}} \over {{A_2}}} = \sqrt {{{0.9 + 0.124} \over {0.9}}} $
$ = \sqrt {{{1.024} \over {0.9}}} $
$ = {\alpha \over {\alpha - 1}}$
$\alpha = 16$
As per given figures, two springs of spring constants $k$ and $2 k$ are connected to mass $m$. If the period of oscillation in figure (a) is $3 \mathrm{s}$, then the period of oscillation in figure (b) will be $\sqrt{x}~ s$. The value of $x$ is ___________.
Explanation:
For case (a),
${K_{eq}} = {{2K} \over 3}$
For case (b),
${K_{eq}} = 3K$
$\because$ $T = 2\pi \sqrt {{m \over K}} $
$\therefore$ ${{{T_a}} \over {{T_b}}} = \sqrt {{{{K_b}} \over {{K_a}}}} $
${3 \over {{T_b}}} = \sqrt {{{3K \times 3} \over {2K}}} = {3 \over {\sqrt 2 }}$
${T_b} = \sqrt 2 $
$x = 2$
A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air jet when it is at 5 cm from its mean position. The new amplitude of vibration is $\sqrt{x}$ cm. The value of x is _____________.
Explanation:
$v = \omega \sqrt {{A^2} - {y^2}} $
$ \Rightarrow 3\omega \sqrt {{{10}^2} - {5^2}} = \omega \sqrt {{{(A')}^2} - {5^2}} $
$ \Rightarrow 9 \times 75 = {(A')^2} - 25$
$ \Rightarrow A' = \sqrt {28 \times 25} $ cm
$ \Rightarrow x = 700$
A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at 60$^\circ$ with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ____________ ms$-$1. (if g = 10 m/s2)
Explanation:
${1 \over 2}m{v^2} = mgl(1 - \cos \theta )$
$ \Rightarrow v = \sqrt {2gl(1 - \cos \theta )} $
$ = \sqrt {2 \times 10 \times 2.5 \times {1 \over 2}} $
$ = 5$ m/s
A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6 s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is ___________ s.
Explanation:
A = 8 cm
T = 6 s
$A\cos \left( {{{2\pi t} \over T}} \right) = {A \over 2}$
$ \Rightarrow {{2\pi t} \over T} = {\pi \over 3}$
or $t = {T \over 6} = 1\,s$
Explanation:
KE = PE
$y = {A \over {\sqrt 2 }} = A\sin \omega t$
$t = {T \over 8} = {T \over x}$
x = 8
Explanation:
${y_2} = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)$
${y_2} = 10\left( {{1 \over 2}\sin 3\pi t + {{\sqrt 3 } \over 2}\cos 3\pi t} \right)$
${y_2} = 10\left( {\cos {\pi \over 3}\sin 3\pi t + \sin {\pi \over 3}\cos 3\pi t} \right)$
${y_2} = 10\left( {3\pi t + {\pi \over 3}} \right)$ $\Rightarrow$ Amplitude = 10
So ratio of amplitudes = ${{10} \over {10}}$ = 1
${x_1} = 5\sin \left( {2\pi t + {\pi \over 4}} \right)$ and ${x_2} = 5\sqrt 2 (\sin 2\pi t + \cos 2\pi t)$. The amplitude of second motion is ................ times the amplitude in first motion.
Explanation:
$ = 10\sin \left( {2\pi t + {\pi \over 4}} \right)$
$\therefore$ ${{{A_2}} \over {{A_1}}} = {{10} \over 5} = 2$
x(t) = A sin($\omega$t + $\phi$)
If the position and velocity of the particle at t = 0 s are 2 cm and 2$\omega$ cm s$-$1 respectively, then its amplitude is $x\sqrt 2 $ cm where the value of x is _________________.
Explanation:
v(t) = A$\omega$ cos ($\omega$t + $\phi$)
2 = A sin$\phi$ ...... (1)
2$\omega$ = A$\omega$ cos$\phi$ ....... (2)
From (1) and (2)
tan$\phi$ = 1
$\phi$ = 45$^\circ$
Putting value of $\phi$ in equation (1),
$2 = A\left\{ {{1 \over {\sqrt 2 }}} \right\}$
$A = 2\sqrt 2 $
$ \therefore $ x = 2
Explanation:
$\mu$ = reduced mass
springs are in series connection
${k_{eq}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}}$
${k_{eq}} = {{k \times 4k} \over {5k}} = {{4k} \over 5}$
${k_{eq}} = {{4 \times 20} \over 5}$ N/m = 16 N/m
$\mu = {{{m_1}{m_2}} \over {{m_1} + {m_2}}} = {{0.2 \times 0.8} \over {0.2 + 0.8}} = 0.16$ kg
$\omega = \sqrt {{{16} \over {0.16}}} = \sqrt {100} = 10$
Explanation:
Applying work energy theorem :
wg + wT = $\Delta$K
$-$mgl(1 $-$ cos60$^\circ$) = ${1 \over 2}$mv2 $-$ ${1 \over 2}$mu2
v2 = u2 $-$ 2gl(1 $-$ cos60$^\circ$)
v2 = 9 $-$ 2 $\times$ 10 $\times$ 0.5$\left( {{1 \over 2}} \right)$
v2 = 4
v = 2 m/s
Explanation:
X = A sin ($\omega$t + $\phi$) ($\phi$ = 0 at M.P.)
$ \Rightarrow $ ${A \over 2} = A\sin {{2\pi } \over T}t$
$ \Rightarrow $ ${{2\pi } \over 2}t = {\pi \over 6}$
$ \Rightarrow $ $t = {1 \over 6}$
$ \therefore $ a = 6
Explanation:
${K_{e{q_{series}}}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}} = {k \over 2}$
${T_b} = 2\pi \sqrt {{M \over {k/2}}} = 2\pi \sqrt {{{2M} \over k}} $
${T_b} = \sqrt 2 {T_a}$
${{{T_b}} \over {{T_a}}} = \sqrt 2 $
$ \therefore $ x = 2
Explanation:
$V = \omega \sqrt {{a^2} - {x^2}} $
Given, $V = {{{V_{\max }}} \over 2} = {{A\omega } \over 2}$
${{{A^2}{\omega ^2}} \over 4} = {\omega ^2}{a^2} - {\omega ^2}{x^2}$
$x = {{\sqrt 3 } \over 2}a$
Explanation:
${5 \over 8}$ oscillation = ${1 \over 2}$ oscillation + ${1 \over 8}$ oscillation
From figure, A to B = ${1 \over 2}$ oscillation and B to C is ${1 \over 8}$ oscillation.
$ \therefore $ $\pi { + \,\theta = \omega t} $
$ \Rightarrow $ $\pi { + {{\pi{} } \over 6}} = \omega t$
$ \Rightarrow $ ${{7\pi {} } \over 6} = \left( {{{2\pi {} } \over T}} \right)t$
$ \Rightarrow $ t = ${{7T} \over {12}}$
Explanation:
Here, we will use the formula for apparent weight in elevator as the elevator has some acceleration.
The normal force (apparent weight) adjusts for the elevations acceleration
relative to gravity.
Apparent weight: $w=m(g+a)$, where $a$ is the elevator's acceleration.
Here, elevator's position, $y$ is given so we can find elevator's acceleration using
$ a=\frac{d^2 y}{d t^2} $
Given, $m=50 \mathrm{~kg}$
$ \begin{aligned} & \text { and } y=8\left[1+\sin \left(\frac{2 \pi t}{T}\right)\right] \\ & T=40 \pi \mathrm{~s} . \\ & \Rightarrow y=8(1+\sin (\omega t)) \quad\left(as, \omega=\frac{2 \pi}{T}\right) \\ & \omega=\frac{2 \pi}{40 \pi}=\frac{1}{20} \\ & a=\frac{d^2 y}{d t^2}=-8 \omega^2 \sin (\omega t) \\ & \left|a_{\text {max }}\right|=8 \omega^2=8\left(\frac{1}{20}\right)^2=8 \times \frac{1}{400} \\ & \left|a_{\text {max }}\right|=\frac{1}{50} \end{aligned} $
$W_{\text {max }}=m\left(g+a_{\text {max }}\right)$ (when lift goes up)
and. $W_{\text {min }}=m\left(g-a_{\text {max }}\right)$ (when lift goes down)
so
$ \begin{aligned} \Delta w & =w_{\text {max }}-w_{\text {min }}=2 m a_{\text {max }} \\ & =2 \times 50 \times \frac{1}{50} \Rightarrow \Delta w=2 \mathrm{~N} \end{aligned} $
Explanation:
At T t$_0$ = 0
Before collision
After collision
$\begin{aligned} \mathrm{v}_{\mathrm{CM}} & =\frac{\mathrm{m} \cdot \frac{\mathrm{a} \omega}{2}+\mathrm{m} \cdot \mathrm{a} \omega}{\mathrm{m}+\mathrm{m}} \\ \mathrm{v}_{\mathrm{CM}} & =\frac{3 \mathrm{a} \omega}{4} \\ \frac{\mathrm{V}_{\mathrm{CM}}}{\mathrm{a} \omega} & =\frac{3}{4} \\ \frac{\mathrm{V}_{\mathrm{CM}}}{\mathrm{a} \omega} & =0.75 \end{aligned}$
Explanation:
$\mathrm{t}_0=\frac{\pi}{2 \omega}=\frac{\mathrm{T}}{4}$
Particles are at extreme position
After collision
in C-frame
using WET,
$\begin{aligned} & \mathrm{W}_{\text {spring }}=\Delta \mathrm{K} \\ & \frac{1}{2} \mathrm{k}(2 \mathrm{~b})^2-\frac{1}{2} \mathrm{k}(2 \mathrm{a})^2=2 \times \frac{1}{2} \mathrm{~m} \times\left(\frac{\mathrm{a} \omega}{4}\right)^2 \quad(\mathrm{k}=\text { spring constant }) \\ & 4 \mathrm{~kb}^2-4 \mathrm{ka}^2=2 \times \mathrm{m} \times \frac{\mathrm{a}^2}{16} \times \frac{2 \mathrm{k}}{\mathrm{m}} \\ & 4 \mathrm{~b}^2=\frac{17}{4} \mathrm{a}^2 \\ & \frac{4 \mathrm{~b}^2}{\mathrm{a}^2}=4.25 \end{aligned}$
Explanation:
So $a_x=\frac{d^2 x}{d t^2}=-x$
$ \Rightarrow x=A_x \sin \left(\omega t+\phi_x\right) \quad(\omega=1 \mathrm{rad} / \mathrm{s}) $
and $v_x=A_x \omega \cos \left(\omega t+\phi_x\right)$
at $t=0, x=\frac{1}{\sqrt{2}} \mathrm{~m}$ and $v_x=-\sqrt{2} \mathrm{~m} / \mathrm{s}$
So $\frac{1}{\sqrt{2}}=A_x \sin \phi_x$
and $-\sqrt{2}=A_x \cos \phi_x$
$ \begin{aligned} & \Rightarrow \tan \phi_x=-\frac{1}{2} ......(1) \\ & \text { and } A_x=\sqrt{\frac{5}{2}} \mathrm{~m} ......(2) \end{aligned} $
Similarly
$ \begin{aligned} & F_y=-y=m a_y . \\\\ & \Rightarrow \frac{d^2 y}{d t^2}=-y \end{aligned} $
So, $y=A_y \sin \left(\omega \mathrm{t}+\phi_{\mathrm{y}}\right) \quad(\omega=1 \mathrm{rad} / \mathrm{s})$ and
$v_y=A_y \omega \cos \left(\omega t+\phi_y\right)$
at $t=0,y=\sqrt{2} \mathrm{~m}$ and $v_y=\sqrt{2} \mathrm{~m} / \mathrm{s}$
So $\sqrt{2}=A_y \sin \phi$
and $\sqrt{2}=A_y \cos \phi$
$ \Rightarrow \phi=\frac{\pi}{4} \text { and } A_y=2 $
So,
$\left(x v_y-y v_x\right) $
$ =\sqrt{\frac{5}{2}} \sin \left(\omega t+\phi_x\right) \times 2 \cos \left(\omega t+\phi_y\right)-2 \sin \left(\omega t+\phi_y\right) \times \sqrt{\frac{5}{2}} \cos \left(\omega t+\phi_x\right) $
$ =\sqrt{\frac{5}{2}} \times 2\left(\sin \left(\omega t+\phi_x\right) \cos \left(\omega t+\phi_y\right)-\sin \left(\omega t+\phi_y\right) \times \cos \left(\omega t+\phi_x\right)\right. $
$ =\sqrt{10} \sin \left(\phi_x-\phi_y\right)$
$ =\sqrt{10}\left(\sin \phi_x \cos \phi_y-\cos \phi_x \sin \phi_y\right) $
$ =\sqrt{10}\left(\frac{1}{\sqrt{5}} \times \frac{1}{\sqrt{2}}-\left(-\frac{2}{\sqrt{5}}\right) \times \frac{1}{\sqrt{2}}\right) $
$=3$
On a frictionless horizontal plane, a bob of mass $m=0.1 \mathrm{~kg}$ is attached to a spring with natural length $l_{0}=0.1 \mathrm{~m}$. The spring constant is $k_{1}=0.009 \,\mathrm{Nm}^{-1}$ when the length of the spring $l>l_{0}$ and is $k_{2}=0.016 \,\mathrm{Nm}^{-1}$ when $l < l_{0}$. Initially the bob is released from $l=$ $0.15 \mathrm{~m}$. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T=(n \pi) s$, then the integer closest to $n$ is __________.
Explanation:
$ \begin{aligned} & =\mathrm{T}_1+\mathrm{T}_2 \\\\ & =\pi \sqrt{\frac{m}{K_1}}+\pi \sqrt{\frac{m}{K_2}} \\\\ & =\pi \sqrt{\frac{0.1}{0.009}}+\pi \sqrt{\frac{0.1}{0.016}} \\\\ & =\frac{\pi}{0.3}+\frac{\pi}{0.4} \\\\ & =\frac{\pi(0.4+0.3)}{0.12} \\\\ & =\frac{70 \pi}{12} \\\\ & =5.83 \pi \text { seconds } \\\\ & \simeq 6 \pi \text { seconds } \end{aligned} $
Explanation:
When mass m is pulled by a force F, the wire elongation x, length l, cross-sectional area A, and Young's modulus of wire material Y are related by $Y = {{F/A} \over {x/l}}$ i.e.,
$F = (Y\,A/l)x$.
The restoring force by the wire is equal but opposite to F i.e., Fr = $-$F. Apply Newton's second law to get
$m{d^2}x/d{t^2} = - (YA/l)x = - {\omega ^2}x$.
This equation represents SHM with an angular frequency $\omega = \sqrt {YA/(lm)} $. Substitute the values to get $Y = {\omega ^2}lm/A = 4 \times {10^9}$ N/m2.
Explanation:
To find the amplitude of oscillation, let's start by considering the given data and the formulae related to simple harmonic motion (SHM).
The total mechanical energy of a system in SHM is constant and is the sum of kinetic energy (KE) and potential energy (PE) at any point in its motion. We are given:
Kinetic Energy, $ KE = 0.5 \, \text{J} $
Potential Energy, $ PE = 0.4 \, \text{J} $
Therefore, the total energy (E) is:
$ E = KE + PE = 0.5 + 0.4 = 0.9 \, \text{J} $
In SHM, the total mechanical energy can also be expressed as:
$ E = \frac{1}{2} k A^2 $
where $ k $ is the spring constant and $ A $ is the amplitude.
We can also express kinetic and potential energy in SHM in terms of amplitude $ A $, position $ x $, and angular frequency $ \omega $:
Kinetic Energy: $ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) $
Potential Energy: $ PE = \frac{1}{2} m \omega^2 x^2 $
Given mass $ m = 0.2 \, \text{kg} $, position $ x = 0.04 \, \text{m} $.
The angular frequency $ \omega $ is related to the frequency $ f $ by:
$ \omega = 2\pi f $
Given frequency:
$ f = \frac{25}{\pi} \, \text{Hz} $
Thus,
$ \omega = 2\pi \left(\frac{25}{\pi}\right) = 50 \, \text{rad/s} $
Using the expression for the total energy in SHM:
$ \frac{1}{2} k A^2 = \frac{1}{2} m \omega^2 A^2 $
Since the total energy $ E = 0.9 \, \text{J} $, it follows:
$ 0.9 = \frac{1}{2} \cdot 0.2 \cdot (50)^2 \cdot A^2 $
Solving for $ A^2 $:
$ 0.9 = \frac{1}{2} \cdot 0.2 \cdot 2500 \cdot A^2 $
$ 0.9 = \frac{1}{2} \cdot 500 \cdot A^2 $
$ 0.9 = 250 \cdot A^2 $
$ A^2 = \frac{0.9}{250} $
$ A^2 = 0.0036 $
$ A = \sqrt{0.0036} $
$ A = 0.06 \, \text{m} $
The amplitude of the oscillations is 0.06 m.