Correct answer is 22

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.

Correct answer is 17

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: $x=A\cos (\omega t+\varphi )$

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

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

Here, $A$ is the amplitude of the motion, $\omega$ is the angular frequency, and $\varphi$ 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:

$a=-A{\omega }^2\cos (\omega t+\varphi )$

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

$16=A{\omega }^2\cos (\omega t+\varphi )$

Using the position equation:

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

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

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

From these two equations, we know:

$A{\omega }^2\cos (\omega t+\varphi )=16$

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

Therefore:

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

$4{\omega }^2=16$

${\omega }^2=4$

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

Next, using the velocity equation:

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

Again, we consider the magnitude:

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

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

$\sin (\omega t+\varphi )={\displaystyle \frac{1}{A}}$

We know from the position equation that:

$\cos (\omega t+\varphi )={\displaystyle \frac{4}{A}}$

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

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

${\displaystyle \frac{1}{A^2}}+{\displaystyle \frac{16}{A^2}}=1$

${\displaystyle \frac{17}{A^2}}=1$

$A^2=17$

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

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

Correct answer is 12

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 $v_{max}$:

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

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

$v_{max}=0.06\times 2$

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

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

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

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

Correct answer is 10

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

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

Where:

$A$ is the amplitude of the SHM,

$\omega$ is the angular frequency,

$t$ is the time,

$\varphi$ 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 $\varphi =\frac{\pi }{3}$. 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.

Correct answer is 7

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 $v^{\prime }$ becomes:

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

For the new amplitude $A^{\prime }$, the velocity $v^{\prime }$ at the same displacement $x$ is:

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

Setting the expressions for $v^{\prime }$ equal gives:

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

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

Solving for $A^{\prime }$ gives:

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

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

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