JEE PYQs Integer Type

Hint

(b) To find the value of \(\beta\), we need to determine the time \(t\) at which the potential energy of the simple harmonic oscillator reaches its maximum value.
Step 1: Understanding Potential Energy in SHM
The potential energy \(U\) of a particle executing simple harmonic motion (SHM) is given by the formula:
\(
U=\frac{1}{2} k x^2
\)
Where:
\(k\) is the force constant (\(k=m \omega^2\))
\(x\) is the displacement at time \(t\)
Given the displacement equation \(x(t)=A \sin \omega t\), we can substitute this into the potential energy formula:
\(
U(t)=\frac{1}{2} k(A \sin \omega t)^2=\frac{1}{2} k A^2 \sin ^2 \omega t
\)
Step 2: Finding the Maximum Potential Energy
The potential energy \(U\) is at its maximum when \(\sin ^2 \omega t=1\). This occurs when:
\(
\sin \omega t= \pm 1
\)
The first instance (after \(t=0\)) where this happens is when the phase angle is \(\pi / 2\) :
\(
\omega t=\frac{\pi}{2}
\)
Step 3: Solving for \(t\) in terms of \(T\)
We know the relationship between angular frequency \(\omega\) and the time period \(T\) is:
\(
\omega=\frac{2 \pi}{T}
\)
Substituting this into our equation for \(t\) :
\(
\begin{gathered}
\left(\frac{2 \pi}{T}\right) t=\frac{\pi}{2} \\
t=\frac{\pi}{2} \cdot \frac{T}{2 \pi} \\
t=\frac{T}{4}
\end{gathered}
\)
Step 4: Comparing with the Given Expression
The problem states that the maximum potential energy is found at:
\(
t=\frac{T}{2 \beta}
\)
By comparing \(t=\frac{T}{4}\) with \(t=\frac{T}{2 \beta}\), we get:
\(
\begin{gathered}
2 \beta=4 \\
\beta=2
\end{gathered}
\)

Hint

(c) Step 1: Identify given parameters
The particle is executing simple harmonic motion (SHM). We are given:
Mass of the particle, \(m=0.50 \mathrm{~kg}\)
Restoring force, \(F=-50 \mathrm{Nm}^{-1} x\)
By comparing this with the standard SHM force equation \(F=-k x\), we find the force constant:
\(
k=50 \mathrm{~N} \mathrm{~m}^{-1}
\)
Step 2: Calculate the time period \(T\)
The formula for the time period \(T\) of a simple harmonic oscillator is:
\(
T=2 \pi \sqrt{\frac{m}{k}}
\)
Substituting the given values and using \(\pi=\frac{22}{7}\) :
\(
\begin{gathered}
T=2 \times \frac{22}{7} \times \sqrt{\frac{0.50}{50}} \\
T=\frac{44}{7} \times \sqrt{\frac{1}{100}} \\
T=\frac{44}{7} \times \frac{1}{10}=\frac{4.4}{7}=\frac{44}{70}=\frac{22}{35} \mathrm{~s}
\end{gathered}
\)
Step 3: Solve for \(\boldsymbol{x}\)
The problem states that the time period is \(\frac{x}{35} \mathrm{~s}\). We equate this to our calculated value:
\(
\frac{x}{35}=\frac{22}{35}
\)
Multiplying both sides by 35 :
\(
x=22
\)

Hint

(d) Step 1: Identify given parameters
The magnitudes of position (\(x\)), velocity (\(v\)), and acceleration (\(a\)) at a specific instant are given as:
Position \((x)=4 \mathrm{~m}\)
Velocity \((v)=2 \mathrm{~ms}^{-1}\)
Acceleration \((a)=16 \mathrm{~ms}^{-2}\)
Step 2: Determine angular frequency (\(\omega\))
In simple harmonic motion (SHM), the magnitude of acceleration is related to displacement by the formula \(a=\omega^2 x\). Using the given values:
\(
\begin{aligned}
16 & =\omega^2 \times 4 \\
\omega^2 & =\frac{16}{4}=4 \\
\omega & =2 \mathrm{rad} / \mathrm{s}
\end{aligned}
\)
Step 3: Calculate the amplitude (\(A\))
The velocity of a particle in SHM is given by the formula \(v=\omega \sqrt{A^2-x^2}\). Squaring both sides gives:
\(
v^2=\omega^2\left(A^2-x^2\right)
\)
Substitute the known values \(\left(v=2, \omega^2=4, x=4\right)\) :
\(
\begin{gathered}
2^2=4\left(A^2-4^2\right) \\
4=4\left(A^2-16\right) \\
1=A^2-16 \\
A^2=17 \\
A=\sqrt{17} \mathrm{~m}
\end{gathered}
\)
Comparing this with the given form \(A=\sqrt{x} \mathrm{~m}\), we find that \(x=17\).

Hint

(c) To find the amplitude in centimeters, we can use the principle of conservation of energy in simple harmonic motion.
Step 1: Calculate Total Mechanical Energy (\(E\))
The total energy of an oscillator is the sum of its kinetic energy (\(K\)) and potential energy (\(U\)) at any given point.
\(
E=K+U
\)
Given:
\(K=0.5 \mathrm{~J}\)
\(U=0.4 \mathrm{~J}\)
\(
E=0.5+0.4=0.9 \mathrm{~J}
\)
Step 2: Determine the Angular Frequency (\(\omega\))
The relationship between frequency (\(f\)) and angular frequency is \(\omega=2 \pi f\).
Given:
\(
f=\frac{25}{\pi} \mathrm{~Hz}
\)
\(
\omega=2 \pi\left(\frac{25}{\pi}\right)=50 \mathrm{rad} / \mathrm{s}
\)
Step 3: Use the Total Energy Formula to find Amplitude (\(\boldsymbol{A}\))
The total energy is also expressed in terms of amplitude as:
\(
E=\frac{1}{2} m \omega^2 A^2
\)
Given:
\(m=0.2 \mathrm{~kg}\)
\(\omega=50 \mathrm{rad} / \mathrm{s}\)
\(E=0.9 \mathrm{~J}\)
Substituting the values:
\(
\begin{gathered}
0.9=\frac{1}{2}(0.2)(50)^2 A^2 \\
0.9=(0.1)(2500) A^2 \\
0.9=250 A^2
\end{gathered}
\)
Step 4: Solve for \(A\)
\(
A^2=\frac{0.9}{250}=\frac{9}{2500}
\)
Taking the square root of both sides:
\(
A=\sqrt{\frac{9}{2500}}=\frac{3}{50} \text { meters }
\)
Step 5: Convert to Centimeters
To convert meters to centimeters, multiply by 100:
\(
A=\frac{3}{50} \times 100=6 \mathrm{~cm}
\)
The amplitude of oscillation is 6 cm.

Hint

(d) To find the maximum velocity of the particle, we can use the standard relationship between amplitude, time period, and angular frequency in SHM.
Step 1: Identify the given values
Amplitude \((A)\) : 0.06 m (which is 6 cm)
Time Period (\(T\)): 3.14 s
Note: Since \(\pi \approx 3.14\), we can treat \(T\) as \(\pi\) seconds to simplify the calculation.
Step 2: Calculate the Angular Frequency (\(\omega\))
The formula for angular frequency is:
\(
\omega=\frac{2 \pi}{T}
\)
Substituting the given time period \((T=3.14 \approx \pi)\) :
\(
\omega=\frac{2 \pi}{\pi}=2 \mathrm{rad} / \mathrm{s}
\)
Step 3: Use the Maximum Velocity Formula
The velocity of a particle in SHM is maximum when it passes through the mean position. The formula for maximum velocity (\(v_{\max }\)) is:
\(
v_{\max }=A \omega
\)
Step 4: Calculate the final value in cm/s
Since the question asks for the answer in \(\mathbf{c m} / \mathbf{s}\), let’s use the amplitude in centimeters (\(A=\) 6 cm):
\(
\begin{gathered}
v_{\max }=6 \mathrm{~cm} \times 2 \mathrm{rad} / \mathrm{s} \\
v_{\max }=12 \mathrm{~cm} / \mathrm{s}
\end{gathered}
\)
The maximum velocity of the particle is \(12 \mathrm{~cm} / \mathrm{s}\).

Hint

(d) To find the velocity of the particle at \(t=0\), we need to derive the velocity equation from the given displacement and use the relationship between the time period and angular frequency.
Step 1: Identify the Displacement Equation
The given displacement is:
\(
x(t)=10 \sin \left(\omega t+\frac{\pi}{3}\right)
\)
From this, we can see the amplitude \(A=10 \mathrm{~m}\) and the initial phase \(\phi=\pi / 3\).
Step 2: Calculate the Angular Frequency (\(\omega\))
The formula for angular frequency is \(\omega=\frac{2 \pi}{T}\).
Given the time period \(T=3.14 \mathrm{~s}\), and noting that \(\pi \approx 3.14\) :
\(
\omega=\frac{2 \pi}{3.14} \approx \frac{2 \pi}{\pi}=2 \mathrm{rad} / \mathrm{s}
\)
Step 3: Derive the Velocity Equation
Velocity \(v(t)\) is the first derivative of displacement with respect to time \(\left(v=\frac{d x}{d t}\right)\) :
\(
\begin{gathered}
v(t)=\frac{d}{d t}\left[10 \sin \left(\omega t+\frac{\pi}{3}\right)\right] \\
v(t)=10 \omega \cos \left(\omega t+\frac{\pi}{3}\right)
\end{gathered}
\)
Step 4: Calculate Velocity at \(t=0\)
Substitute \(t=0\) and \(\omega=2 \mathrm{rad} / \mathrm{s}\) into the velocity equation:
\(
\begin{gathered}
v(0)=10(2) \cos \left(2(0)+\frac{\pi}{3}\right) \\
v(0)=20 \cos \left(\frac{\pi}{3}\right)
\end{gathered}
\)
Since \(\cos \left(\frac{\pi}{3}\right)=\frac{1}{2}\) :
\(
v(0)=20 \times \frac{1}{2}=10 \mathrm{~m} / \mathrm{s}
\)
The velocity of the particle at \(t=0\) is \(10 \mathrm{~m} / \mathrm{s}\).

Hint

(c) To find the ratio of the frequencies, we need to look at how the mass of an oscillating system affects its frequency of vibration.
Step 1: Identify the Frequency Formula
For a mass-spring system, the frequency (\(f\)) is given by the formula:
\(
f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}
\)
Where:
\(k\) is the spring constant (which remains the same since it’s the “same spring”).
\(m\) is the suspended mass.
Step 2: Set up the expressions for \(f_1\) and \(f_2\)
Case 1: For mass \(m\), the frequency is \(f_1\) :
\(
f_1=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}
\)
Case 2: For mass \(9 m\), the frequency is \(f_2\) :
\(
f_2=\frac{1}{2 \pi} \sqrt{\frac{k}{9 m}}
\)
Step 3: Simplify the expression for \(f_2\)
We can take the constant \(\sqrt{9}\) out of the square root:
\(
f_2=\frac{1}{3}\left(\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\right)
\)
Notice that the term in the parentheses is exactly \(f_1\). So:
\(
f_2=\frac{f_1}{3}
\)
Step 4: Calculate the ratio \(\frac{f_1}{f_2}\)
By rearranging the equation from Step 3:
\(
\frac{f_1}{f_2}=3
\)
The value of \(\frac{f_1}{f_2}\) is 3.

Hint

(a) This configuration is a classic “mixed” spring system. To find the value of \(\alpha\), we need to calculate the equivalent spring constant \(\left(k_{\text {eq }}\right)\) step-by-step.
Step 1: Calculate the Parallel Combination
On the right side, the two upper springs are connected in parallel because they share the same displacement.
\(
k_{\text {parallel }}=k+k=2 k
\)
Step 2: Calculate the Series Combination
This \(2 k\) combination is connected in series with the spring directly below it. The equivalent constant for this right-hand branch (\(k_{\text {right }}\)) is:
\(
\begin{gathered}
\frac{1}{k_{\text {right }}}=\frac{1}{2 k}+\frac{1}{k} \\
\frac{1}{k_{\text {right }}}=\frac{1+2}{2 k}=\frac{3}{2 k} \Longrightarrow k_{\text {right }}=\frac{2 k}{3}
\end{gathered}
\)
Step 3: Calculate the Total Equivalent Spring Constant (\(k_{\text {eq }}\))
Now, the single spring on the left (\(k\)) and the entire right branch (\(k_{\text {right }}\)) are connected in parallel to the mass \(M\) (if the mass moves down by \(x\), both sides stretch by \(x\)).
\(
\begin{gathered}
k_{\text {eq }}=k+k_{\text {right }} \\
k_{\text {eq }}=k+\frac{2 k}{3}=\frac{3 k+2 k}{3}=\frac{5 k}{3}
\end{gathered}
\)
Step 4: Find the Time Period and Solve for \(\alpha\)
The standard formula for the time period is:
\(
T=2 \pi \sqrt{\frac{M}{k_{e q}}}
\)
Substitute \(k_{\text {eq }}=\frac{5 k}{3}\) :
\(
T=2 \pi \sqrt{\frac{M}{5 k / 3}}=2 \pi \sqrt{\frac{3 M}{5 k}}
\)
To match your expression \(\pi \sqrt{\frac{\alpha M}{5 k}}\), we need to move the 2 inside the square root \((2=\sqrt{4})\) :
\(
T=\pi \sqrt{4 \cdot \frac{3 M}{5 k}}=\pi \sqrt{\frac{12 M}{5 k}}
\)
Comparing this to the given expression:
\(
\alpha=12
\)

Hint

(d) To solve for the new amplitude, we need to find the velocity of the particle at the given displacement and then apply the energy conservation principle for the “new” state of motion.
Step 1: Find the initial velocity (\(v\)) at displacement \(x=\frac{2 A}{3}\)
The velocity of a particle in SHM is given by:
\(
v=\omega \sqrt{A^2-x^2}
\)
Substituting \(x=\frac{2 A}{3}\) :
\(
\begin{gathered}
v=\omega \sqrt{A^2-\left(\frac{2 A}{3}\right)^2} \\
v=\omega \sqrt{A^2-\frac{4 A^2}{9}}=\omega \sqrt{\frac{5 A^2}{9}} \\
v=\frac{\omega A \sqrt{5}}{3}
\end{gathered}
\)
Step 2: Determine the new velocity (\(v^{\prime}\))
The problem states the speed is increased to three times its current value:
\(
\begin{gathered}
v^{\prime}=3 v=3\left(\frac{\omega A \sqrt{5}}{3}\right) \\
v^{\prime}=\omega A \sqrt{5}
\end{gathered}
\)
Step 3: Calculate the new amplitude (\(\boldsymbol{A}^{\boldsymbol{\prime}}\))
Even though the velocity changed, the angular frequency \(\omega\) (which depends on the spring and mass) remains the same. We use the velocity-displacement formula again for the new amplitude \(A^{\prime}\) at the same position \(x=\frac{2 A}{3}\) :
\(
\left(v^{\prime}\right)^2=\omega^2\left(A^{\prime 2}-x^2\right)
\)
Substitute \(v^{\prime}=\omega A \sqrt{5}\) and \(x=\frac{2 A}{3}\) :
\(
\begin{aligned}
(\omega A \sqrt{5})^2 & =\omega^2\left(A^{\prime 2}-\left(\frac{2 A}{3}\right)^2\right) \\
5 \omega^2 A^2 & =\omega^2\left(A^{\prime 2}-\frac{4 A^2}{9}\right)
\end{aligned}
\)
Divide both sides by \(\omega^2\) :
\(
\begin{gathered}
5 A^2=A^{\prime 2}-\frac{4 A^2}{9} \\
A^{\prime 2}=5 A^2+\frac{4 A^2}{9} \\
A^{\prime 2}=\frac{45 A^2+4 A^2}{9}=\frac{49 A^2}{9}
\end{gathered}
\)
Taking the square root:
\(
A^{\prime}=\frac{7 A}{3}
\)
Step 4: Compare and find \(n\)
The problem defines the new amplitude as \(\frac{n A}{3}\). Comparing this to our result:
\(
\begin{aligned}
\frac{n A}{3} & =\frac{7 A}{3} \\
n & =7
\end{aligned}
\)

Hint

(b) To find the value of \(x\), we need to determine the time interval between two specific positions of the particle.
Step 1: Identify the Equation of Motion
Since the oscillation starts from the mean position at \(t=0\), the displacement equation is:
\(
x(t)=A \sin (\omega t)
\)
Step 2: Calculate the Angular Frequency (\(\omega\))
Given the time period \(T=6 \pi \mathrm{~s}\) :
\(
\omega=\frac{2 \pi}{T}=\frac{2 \pi}{6 \pi}=\frac{1}{3} \mathrm{rad} / \mathrm{s}
\)
Step 3: Find the time to reach \(x=A\)
The particle reaches its amplitude \(A\) (the extreme position) when the phase is \(\pi / 2\) :
\(
\begin{gathered}
\omega t_1=\frac{\pi}{2} \\
\frac{1}{3} t_1=\frac{\pi}{2} \Longrightarrow t_1=\frac{3 \pi}{2} \mathrm{~s}
\end{gathered}
\)
Step 4: Find the time to reach \(x=\frac{\sqrt{3}}{2} A\)
Set the displacement equation to the target value:
\(
\begin{gathered}
\frac{\sqrt{3}}{2} A=A \sin \left(\omega t_2\right) \\
\sin \left(\omega t_2\right)=\frac{\sqrt{3}}{2}
\end{gathered}
\)
The first instance this occurs is at \(\omega t_2=\frac{\pi}{3}\) :
\(
\frac{1}{3} t_2=\frac{\pi}{3} \Longrightarrow t_2=\pi \mathrm{s}
\)
Step 5: Calculate the Time Interval
The time required to travel from \(x=A\) to \(x=\frac{\sqrt{3}}{2} A\) is the difference between these two times:
\(
\begin{gathered}
\Delta t=t_1-t_2=\frac{3 \pi}{2}-\pi \\
\Delta t=\frac{\pi}{2} \mathrm{~s}
\end{gathered}
\)
Step 6: Solve for \(x\)
The problem states the time required is \(\frac{\pi}{x} \mathrm{~s}\). Comparing this to our result:
\(
\begin{aligned}
\frac{\pi}{x} & =\frac{\pi}{2} \\
x & =2
\end{aligned}
\)

Hint

(c) To find the value of \(x\), we need to express both the total energy and the kinetic energy in terms of the amplitude and displacement.
Step 1: Write the expression for Total Energy (\(E\))
The total mechanical energy of a simple harmonic oscillator is constant and depends only on the amplitude \(A\) :
\(
E=\frac{1}{2} k A^2
\)
Step 2: Write the expression for Kinetic Energy (\(K\))
The kinetic energy at any displacement \(y\) is the total energy minus the potential energy (\(U= \left.\frac{1}{2} k y^2\right)\) :
\(
\begin{gathered}
K=E-U=\frac{1}{2} k A^2-\frac{1}{2} k y^2 \\
K=\frac{1}{2} k\left(A^2-y^2\right)
\end{gathered}
\)
Step 3: Substitute the given displacement
The problem states the displacement is one-third of the amplitude \(\left(y=\frac{A}{3}\right)\). Substituting this into the kinetic energy formula:
\(
\begin{gathered}
K=\frac{1}{2} k\left(A^2-\left(\frac{A}{3}\right)^2\right) \\
K=\frac{1}{2} k\left(A^2-\frac{A^2}{9}\right) \\
K=\frac{1}{2} k\left(\frac{8 A^2}{9}\right)
\end{gathered}
\)
Step 4: Calculate the ratio of Total Energy to Kinetic Energy
Now, we divide the Total Energy (\(E\)) by the Kinetic Energy (\(K\)):
\(
\frac{E}{K}=\frac{\frac{1}{2} k A^2}{\frac{1}{2} k\left(\frac{8 A^2}{9}\right)}
\)
The terms \(\frac{1}{2} k A^2\) cancel out, leaving:
\(
\frac{E}{K}=\frac{1}{8 / 9}=\frac{9}{8}
\)
Step 5: Solve for \(x\)
The problem gives this ratio as \(\frac{x}{8}\). Comparing our result to the given expression:
\(
\begin{aligned}
& \frac{9}{8}=\frac{x}{8} \\
& x=9
\end{aligned}
\)

Hint

(c) To find the value of \(\alpha\), we need to use the relationship between velocity, amplitude, and displacement in simple harmonic motion (SHM).
Step 1: Find the angular frequency (\(\omega\))
In SHM, the velocity of the particle at the mean position is the maximum velocity (\(v_{\text {max }}\)). The formula is:
\(
v_{\max }=A \omega
\)
Given:
\(A=4 \mathrm{~cm}\)
\(v_{\max }=10 \mathrm{~cm} / \mathrm{s}\)
Substituting the values:
\(
10=4 \omega \Longrightarrow \omega=\frac{10}{4}=2.5 \mathrm{rad} / \mathrm{s}
\)
Step 2: Use the velocity-displacement formula
The velocity (\(v\)) at any distance (\(y\)) from the mean position is given by:
\(
v=\omega \sqrt{A^2-y^2}
\)
Given:
\(v=5 \mathrm{~cm} / \mathrm{s}\)
\(\omega=2.5 \mathrm{rad} / \mathrm{s}\)
\(A=4 \mathrm{~cm}\)
Step 3: Solve for the distance (\(y\))
Substitute the values into the equation:
\(
5=2.5 \sqrt{4^2-y^2}
\)
Divide both sides by 2.5:
\(
2=\sqrt{16-y^2}
\)
\(
y=\sqrt{12} \mathrm{~cm}
\)
Step 4: Compare and find \(\alpha\)
The problem states the distance is \(\sqrt{\alpha} \mathrm{cm}\). Comparing this to our result:
\(
\begin{gathered}
\sqrt{\alpha}=\sqrt{12} \\
\alpha=12
\end{gathered}
\)

Hint

(d) Step 1: Analyze the Displacement (\(y\))
The problem gives the displacement as a function of the amplitude \(A\) and a phase angle.
\(
y=A \cos \left(30^{\circ}\right)
\)
Using the trigonometric value \(\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}\) :
\(
y=A\left(\frac{\sqrt{3}}{2}\right)
\)
Given \(A=40 \mathrm{~cm}=0.4 \mathrm{~m}\), we find:
\(
y=0.4 \times \frac{\sqrt{3}}{2}=0.2 \sqrt{3} \mathrm{~m}
\)
Step 2: Use the Kinetic Energy Formula
The kinetic energy (\(K\)) of an SHO at any displacement \(y\) is the difference between the total energy and the potential energy:
\(
K=\frac{1}{2} k\left(A^2-y^2\right)
\)
Step 3: Substitute and Solve for \(k\)
Substitute the known values (\(K=200 \mathrm{~J}, A=0.4\), and \(y=0.2 \sqrt{3}\)):
\(
200=\frac{1}{2} k\left[(0.4)^2-(0.2 \sqrt{3})^2\right]
\)
\(k=\frac{200}{0.02}=10,000 \mathrm{~N} / \mathrm{m}\)
Step 4: Find the Value of \(x\)
Express the force constant in scientific notation:
\(
k=1.0 \times 10^4 \mathrm{~N} / \mathrm{m}
\)
Comparing this to the required form \(1.0 \times 10^x\), we get:
\(
x=4
\)

Hint

(b) Step 1: Calculate the spring constant (\(k\))
The time period \(T\) of a mass-spring system is related to the mass \(m\) and the spring constant \(k\) by the formula:
\(
T=2 \pi \sqrt{\frac{m}{k}}
\)
Rearranging this formula to solve for \(k\) :
\(
T^2=4 \pi^2 \frac{m}{k} \Longrightarrow k=\frac{4 \pi^2 m}{T^2}
\)
Given \(T=3.14 \mathrm{~s}\) (which is approximately \(\pi\)) and \(m=5 \mathrm{~kg}\) :
\(
k=\frac{4 \pi^2 \times 5}{(3.14)^2}
\)
Since \(3.14^2 \approx \pi^2\), they cancel out:
\(
k \approx 4 \times 5=20 \mathrm{~N} / \mathrm{m}
\)
Step 2: Calculate the maximum force (\(F_{\text {max }}\))
According to Hooke’s Law, the force exerted by the spring is \(F=k x\). The force is at its maximum when the displacement \(x\) is equal to the amplitude \(A\).
\(
F_{\max }=k A
\)
Substituting the values \(k=20 \mathrm{~N} / \mathrm{m}\) and \(A=1 \mathrm{~m}\) :
\(
F_{\max }=20 \times 1=20 \mathrm{~N}
\)
The maximum force exerted by the spring on the block is 20 N.

Hint

(c) To find the value of \(x\), we need to calculate the maximum tension in the string during the pendulum’s oscillation.
Step 1: Identify Given Values
Length \((L): 100 \mathrm{~cm}=1.0 \mathrm{~m}\)
Mass (\(m\)): \(250 \mathrm{~g}=0.25 \mathrm{~kg}\)
Amplitude (\(A\)): \(10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Gravity (\(g\)): \(10 \mathrm{~m} / \mathrm{s}^2\)
Step 2: Determine where Tension is Maximum
The tension \((T)\) in a simple pendulum is maximum at the mean position (the lowest point of the swing). At this point, the tension must support the weight of the bob and provide the centripetal force required for circular motion.
\(
T_{\max }=m g+\frac{m v_{\max }^2}{L}
\)
Step 3: Calculate Maximum Velocity (\(v_{\text {max }}\))
In SHM, the maximum velocity is given by \(v_{\text {max }}=\omega A\).
First, find the angular frequency \(\omega\) :
\(
\begin{aligned}
&\omega=\sqrt{\frac{g}{L}}=\sqrt{\frac{10}{1}}=\sqrt{10} \mathrm{rad} / \mathrm{s}\\
&\text { Now, calculate } v_{\text {max }} \text { : }\\
&\begin{gathered}
v_{\max }=\sqrt{10} \times 0.1 \mathrm{~m} / \mathrm{s} \\
v_{\max }^2=10 \times 0.01=0.1 \mathrm{~m}^2 / \mathrm{s}^2
\end{gathered}
\end{aligned}
\)
Step 4: Calculate Maximum Tension
Substitute the values into the tension formula:
\(
\begin{gathered}
T_{\max }=(0.25 \times 10)+\frac{0.25 \times 0.1}{1} \\
T_{\max }=2.5+0.025 \\
T_{\max }=2.525 \mathrm{~N}
\end{gathered}
\)
Step 5: Solve for \(x\)
The problem states \(T_{\text {max }}=\frac{x}{40}\).
\(
\begin{gathered}
2.525=\frac{x}{40} \\
x=2.525 \times 40 \\
x=101
\end{gathered}
\)
The value of \(x\) is 101.

Hint

(b) To find the displacement where the kinetic energy \((K)\) is \(25 \%\) more than the potential energy (\(U\)), we follow these steps:
Step 1: Set up the Energy Relationship
The problem states that \(K\) is \(25 \%\) greater than \(U\). This can be written as:
\(
\begin{gathered}
K=U+25 \% \text { of } U \\
K=U+0.25 U \\
K=1.25 U=\frac{5}{4} U
\end{gathered}
\)
Step 2: Use SHM Energy Formulas
In Simple Harmonic Motion, at any displacement \(x\) from the mean position:
Potential Energy: \(U=\frac{1}{2} k x^2\)
Kinetic Energy: \(K=\frac{1}{2} k\left(A^2-x^2\right)\)
Substitute these formulas into our relationship from Step 1:
\(
\frac{1}{2} k\left(A^2-x^2\right)=\frac{5}{4}\left(\frac{1}{2} k x^2\right)
\)
\(
x=\frac{2}{3} A
\)
Step 4: Substitute the Given Amplitude
Given that the amplitude \(A=3 \mathrm{~cm}\) :
\(
\begin{aligned}
& x=\frac{2}{3} \times 3 \\
& x=2 \mathrm{~cm}
\end{aligned}
\)
The displacement at which the kinetic energy is \(25 \%\) more than the potential energy is 2 cm.

Hint

(a) To solve for the number of complete oscillations, we first need to determine the effective spring constant (\(K_{\text {eff }}\)) of the system and then find the time period of oscillation.
Step 1: Identify the Spring Arrangement
In the given figure, the block is connected between two springs attached to fixed walls. When the block is displaced to one side (say, to the right) by a distance \(x\) :
The left spring is stretched by \(x\), pulling the block back to the left.
The right spring is compressed by \(x\), pushing the block back to the left.
Since both springs provide a restoring force in the same direction, they act in parallel.
\(
K_{\mathrm{eff}}=K+K=2 K
\)
Given \(K=2 \mathrm{Nm}^{-1}\), we have:
\(
K_{\mathrm{eff}}=2 \times 2=4 \mathrm{Nm}^{-1}
\)
Step 2: Calculate the Time Period (\(T\))
The time period for a mass-spring system is given by:
\(
T=2 \pi \sqrt{\frac{M}{K_{\mathrm{eff}}}}
\)
Substitute the values \(M=490 \mathrm{~g}=0.49 \mathrm{~kg}\) and \(K_{\text {eff }}=4 \mathrm{Nm}^{-1}\) :
\(
\begin{aligned}
T & =2 \pi \sqrt{\frac{0.49}{4}} \\
T & =2 \pi\left(\frac{0.7}{2}\right) \\
T & =0.7 \pi \text { seconds }
\end{aligned}
\)
Step 3: Calculate the Number of Oscillations (\(n\))
The total time given is \(t=14 \pi\) seconds. The number of complete oscillations is the total time divided by the time for one oscillation:
\(
\begin{gathered}
n=\frac{t}{T} \\
n=\frac{14 \pi}{0.7 \pi} \\
n=\frac{14}{0.7}=20
\end{gathered}
\)
The number of complete oscillations the block will make in \(14 \pi\) seconds is 20.

Note: In the given configuration, the block is connected between two springs fixed to walls. When the block is displaced, one spring is compressed while the other is stretched. Both springs exert a restoring force in the same direction, meaning they act in parallel. The effective spring constant \(k_{\text {eff }}\) is:\(k_{e f f}=K+K=2K\)

Hint

(c) To find the value of \(x\), we need to compare the given velocity equation with the standard equation for a particle executing Simple Harmonic Motion (SHM).
Step 1: Standard Equation of Velocity in SHM
The velocity \(v\) of a particle in SHM at a displacement \(x\) is given by:
\(
v=\omega \sqrt{A^2-x^2}
\)
Squaring both sides gives:
\(
v^2=\omega^2\left(A^2-x^2\right)
\)
Step 2: Rearrange the Given Equation
The problem provides the following relation:
\(
4 v^2=50-x^2
\)
Divide the entire equation by 4 to isolate \(v^2\) :
\(
\begin{aligned}
v^2 & =\frac{50}{4}-\frac{x^2}{4} \\
v^2 & =\frac{1}{4}\left(50-x^2\right)
\end{aligned}
\)
Step 3: Compare and Find Angular Frequency (\(\omega\))
By comparing our rearranged equation \(v^2=\frac{1}{4}\left(50-x^2\right)\) with the standard form \(v^2= \omega^2\left(A^2-x^2\right)\), we can identify the coefficients:
\(
\omega^2=\frac{1}{4}
\)
Taking the square root:
\(
\omega=\frac{1}{2} \mathrm{rad} / \mathrm{s}
\)
Step 4: Calculate the Time Period (\(T\))
The relationship between the time period \(T\) and angular frequency \(\omega\) is:
\(
T=\frac{2 \pi}{\omega}
\)
Substitute \(\omega=1 / 2\) and \(\pi=22 / 7\) :
\(
\begin{gathered}
T=\frac{2 \times(22 / 7)}{1 / 2} \\
T=2 \times \frac{22}{7} \times 2 \\
T=\frac{88}{7} \mathrm{~s}
\end{gathered}
\)
Step 5: Solve for \(x\)
The problem states that the time period is \(\frac{x}{7} \mathrm{~s}\).
\(
\begin{aligned}
& \frac{88}{7}=\frac{x}{7} \\
& x=88
\end{aligned}
\)

Hint

(d) To find the value of \(\beta\), we need to analyze the potential energy function \(U(t)\) and find where its rate of change (slope) is at a maximum.
Step 1: Express Potential Energy (\(U\)) as a function of time (\(t\))
The potential energy of a simple harmonic oscillator is given by:
\(
U=\frac{1}{2} k x^2
\)
Substituting the given displacement \(x=A \sin \omega t\) :
\(
U(t)=\frac{1}{2} k A^2 \sin ^2 \omega t
\)
Using the trigonometric identity \(\sin ^2 \theta=\frac{1-\cos 2 \theta}{2}\), we can rewrite this as:
\(
U(t)=\frac{1}{4} k A^2(1-\cos 2 \omega t)
\)
Step 2: Find the Slope of the \(U-t\) curve
The slope is the first derivative of potential energy with respect to time:
\(
\begin{aligned}
\text { Slope } & =\frac{d U}{d t}=\frac{d}{d t}\left[\frac{1}{4} k A^2(1-\cos 2 \omega t)\right] \\
\frac{d U}{d t} & =\frac{1}{4} k A^2(0-(-\sin 2 \omega t \cdot 2 \omega))
\end{aligned}
\)
\(
\frac{d U}{d t}=\frac{1}{2} k A^2 \omega \sin 2 \omega t
\)
Step 3: Maximize the Slope
The slope is a sine function: Slope \(\propto \sin 2 \omega t\). A sine function reaches its maximum value when its argument is \(90^{\circ}\) (or \(\pi / 2\) radians).
\(
2 \omega t=\frac{\pi}{2}
\)
Step 4: Solve for \(t\) in terms of \(T\)
We know that the angular frequency \(\omega=\frac{2 \pi}{T}\). Substitute this into the equation:
\(
\begin{gathered}
2\left(\frac{2 \pi}{T}\right) t=\frac{\pi}{2} \\
\frac{4 \pi t}{T}=\frac{\pi}{2} \\
4 t=\frac{T}{2} \\
t=\frac{T}{8}
\end{gathered}
\)
Step 5: Identify \(\beta\)
The problem states the time is \(t=\frac{T}{\beta}\). Comparing this to our result \(t=\frac{T}{8}\) :
\(
\beta=8
\)

Hint

(a) To find the amplitude of the motion, we need to relate the force constant, the mass, and the maximum speed of the particle.
Step 1: Find the Force Constant (\(k\))
The restoring force in SHM is given by \(F=-k x\). By comparing this to the given equation:
\(
F=-25 x
\)
We can see that the force constant \(k=25 \mathrm{~N} / \mathrm{m}\).
Step 2: Calculate the Angular Frequency (\(\omega\))
The angular frequency is determined by the mass (\(m\)) and the force constant (\(k\)).
Given \(m=250 \mathrm{~g}=0.25 \mathrm{~kg}\) :
\(
\begin{gathered}
\omega=\sqrt{\frac{k}{m}} \\
\omega=\sqrt{\frac{25}{0.25}}=\sqrt{100} \\
\omega=10 \mathrm{rad} / \mathrm{s}
\end{gathered}
\)
Step 3: Use the Maximum Speed Formula
In SHM, the maximum speed (\(v_{\text {max }}\)) occurs at the mean position and is defined by:
\(
v_{\max }=\omega A
\)
Where \(A\) is the amplitude. Substituting the given \(v_{\max }=4 \mathrm{~m} / \mathrm{s}\) and our calculated \(\omega= 10 \mathrm{rad} / \mathrm{s}\) :
\(
\begin{gathered}
4=10 \times A \\
A=\frac{4}{10}=0.4 \mathrm{~m}
\end{gathered}
\)
Step 4: Convert to Centimeters
The question asks for the answer in cm:
\(
\begin{gathered}
A=0.4 \times 100 \\
A=40 \mathrm{~cm}
\end{gathered}
\)
The amplitude of the motion is 40 cm.

Hint

(a) To find the value of the initial mass \(m\), we can use the relationship between the mass and the time period of a spring-mass system.
Step 1: Set up the initial condition
The time period \(T\) of a spring-mass system is given by the formula:
\(
T=2 \pi \sqrt{\frac{m}{k}}
\)
For the first case, we are given \(T_1=1 \mathrm{~s}\) :
\(
1=2 \pi \sqrt{\frac{m}{k}} \dots(1)
\)
Step 2: Set up the second condition
When the mass is increased by 3 kg, the new mass becomes (\(m+3\)). The problem states the period increases by 1 second, so the new period \(T_2=1+1=2 \mathrm{~s}\).
\(
2=2 \pi \sqrt{\frac{m+3}{k}} \dots(2)
\)
Step 3: Divide the equations to eliminate \(k\)
Divide Eq. 2 by Eq. 1:
\(
\begin{aligned}
& \frac{2}{1}=\frac{2 \pi \sqrt{\frac{m+3}{k}}}{2 \pi \sqrt{\frac{m}{k}}} \\
& 2=\sqrt{\frac{m+3}{m}}
\end{aligned}
\)
Step 4: Solve for \(m\)
Square both sides of the equation to remove the square root:
\(
\begin{gathered}
4=\frac{m+3}{m} \\
4 m=m+3 \\
3 m=3 \\
m=1 \mathrm{~kg}
\end{gathered}
\)
The value of the initial mass \(m\) is 1 kg.

Hint

(d) To find the value of \(x\), we need to determine the effective spring constant for this configuration and then calculate the time period of oscillation.
Step 1: Determine the Effective Spring Constant (\(k_{\text {eff }}\))
In the arrangement shown, the block is connected between two identical springs. When the block is displaced to the right by a distance \(x\) :
The left spring is stretched by \(x\), pulling the block to the left with force \(k x\).
The right spring is compressed by \(x\), pushing the block to the left with force \(k x\).
Since both forces act in the same direction to restore the block to equilibrium, the springs are in a parallel-like arrangement.
\(
k_{\mathrm{eff}}=k+k=2 k
\)
Given \(k=20 \mathrm{~N} / \mathrm{m}\) :
\(
k_{\mathrm{eff}}=2 \times 20=40 \mathrm{~N} / \mathrm{m}
\)
Step 2: Calculate the Time Period (\(T\))
The formula for the time period of a mass-spring system is:
\(
T=2 \pi \sqrt{\frac{m}{k_{\mathrm{eff}}}}
\)
Substitute \(m=2 \mathrm{~kg}\) and \(k_{\text {eff }}=40 \mathrm{~N} / \mathrm{m}\) :
\(
T=\frac{2 \pi}{\sqrt{20}}
\)
Step 3: Simplify and Find \(x\)
We need to bring the expression into the form \(\frac{\pi}{\sqrt{x}}\). Let’s manipulate the expression:
\(
T=\frac{2 \pi}{\sqrt{20}}=\frac{\pi}{\frac{\sqrt{20}}{2}}
\)
To bring the 2 inside the square root, we square it \(\left(2^2=4\right)\) :
\(
\begin{gathered}
T=\frac{\pi}{\sqrt{\frac{20}{4}}} \\
T=\frac{\pi}{\sqrt{5}}
\end{gathered}
\)
Comparing this to the given form \(T=\frac{\pi}{\sqrt{x}}\) :
\(
x=5
\)

Hint

(b)

\(
\mathrm{mg}^{\prime}=\mathrm{mg}-\mathrm{F}_{\mathrm{B}}
\)
\(
\begin{aligned}
& g^{\prime}=\frac{m g-F_B}{m} \\
& =\frac{\rho_B V g-\rho_w V g}{\rho_B V} \\
& =\left(\frac{\rho_B-\rho_w}{\rho_B}\right) g
\end{aligned}
\)

\(
\begin{aligned}
& =\frac{5-1}{5} \times \mathrm{g} \\
& =\frac{4}{5} \mathrm{~g}
\end{aligned}
\)
\(
\mathbf{T}=2 \pi \sqrt{\frac{\ell}{\mathbf{g}}}
\)
\(
\begin{aligned}
& \frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{\mathrm{~g}^{\prime}}}=\sqrt{\frac{\mathrm{g}}{\frac{4}{5} \mathrm{~g}}}=\sqrt{\frac{5}{4}} \\
& \mathrm{~T}^{\prime}=\mathrm{T} \sqrt{\frac{5}{4}}=\frac{10}{2} \sqrt{5} \\
& \mathrm{~T}^{\prime}=5 \sqrt{5}
\end{aligned}
\)

Alternate: To find the value of \(x\), we need to calculate how the “effective gravity” changes when the pendulum bob is immersed in water.
Step 1: Effective Gravity in Water
When the bob is in air, the acceleration acting on it is simply \(g\). When immersed in water, an upward buoyant force acts on the bob, reducing its effective weight.
The effective acceleration \(\left(g^{\prime}\right)\) is given by:
\(
g^{\prime}=g\left(1-\frac{\rho_{\mathrm{water}}}{\rho_{\mathrm{bob}}}\right)
\)
Given the relative density \((\sigma)\) of the bob is 5, we know that \(\frac{\rho_{\text {bob }}}{\rho_{\text {water }}}=5\). Therefore:
\(
g^{\prime}=g\left(1-\frac{1}{5}\right)=g\left(\frac{4}{5}\right)
\)
Step 2: Relation between Time Period and Gravity
The time period \(T\) of a simple pendulum is inversely proportional to the square root of the acceleration due to gravity:
\(
T \propto \frac{1}{\sqrt{g}}
\)
We can set up a ratio between the initial period (\(T=10 \mathrm{~s}\)) and the new period (\(T^{\prime}\)):
\(
\frac{T^{\prime}}{T}=\sqrt{\frac{g}{g^{\prime}}}
\)
Step 3: Calculate the New Time Period
Substitute the value of \(g^{\prime}\) into the ratio:
\(
\begin{gathered}
\frac{T^{\prime}}{10}=\sqrt{\frac{g}{4} g} \\
\frac{T^{\prime}}{10}=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}
\end{gathered}
\)
Now, solve for \(T^{\prime}\) :
\(
\begin{aligned}
T^{\prime} & =10 \times \frac{\sqrt{5}}{2} \\
T^{\prime} & =5 \sqrt{5} \mathrm{~s}
\end{aligned}
\)
Step 4: Find the Value of \(x\)
The problem states the new time period is \(5 \sqrt{x} \mathrm{~s}\). Comparing this to our result \(5 \sqrt{5} \mathrm{~s}\) :
\(
x=5
\)

Hint

(b) To find the value of \(K\), we need to determine the angular frequency \(\omega\) of the system by examining the potential energy function \(U(x)\) for small oscillations.
Step 1: Use the Small Angle Approximation
The potential energy is given as \(U=4(1-\cos 4 x)\). For small oscillations, \(x\) is very small. We can use the Taylor series expansion for \(\cos \theta \approx 1-\frac{\theta^2}{2}\) :
\(
\cos 4 x \approx 1-\frac{(4 x)^2}{2}=1-8 x^2
\)
Substituting this into the potential energy equation:
\(
\begin{gathered}
U \approx 4\left[1-\left(1-8 x^2\right)\right] \\
U \approx 4\left(8 x^2\right)=32 x^2
\end{gathered}
\)
Step 2: Identify the Force Constant (\(k\))
The standard form for potential energy in Simple Harmonic Motion (SHM) is:
\(
U=\frac{1}{2} k x^2
\)
Comparing \(U=32 x^2\) with the standard form:
\(
\frac{1}{2} k=32 \Longrightarrow k=64 \mathrm{~N} / \mathrm{m}
\)
Step 3: Calculate the Angular Frequency (\(\omega\))
The angular frequency is determined by the mass (\(m=4 \mathrm{~kg}\)) and the force constant (\(k= 64 \mathrm{~N} / \mathrm{m}\)):
\(
\begin{gathered}
\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{64}{4}}=\sqrt{16} \\
\omega=4 \mathrm{rad} / \mathrm{s}
\end{gathered}
\)
Step 4: Calculate the Time Period (\(T\))
The relationship between the time period and angular frequency is:
\(
\begin{gathered}
T=\frac{2 \pi}{\omega} \\
T=\frac{2 \pi}{4}=\frac{\pi}{2} \mathrm{~s}
\end{gathered}
\)
Step 5: Solve for \(K\)
The problem states the time period is \(\left(\frac{\pi}{K}\right) \mathrm{s}\). Comparing this to our result:
\(
\begin{gathered}
\frac{\pi}{K}=\frac{\pi}{2} \\
K=2
\end{gathered}
\)

Hint

(d) This problem involves two key physics principles: Conservation of Linear Momentum (at the mean position) and the Conservation of Mechanical Energy (to relate velocity to amplitude).
Step 1: Conservation of Momentum at the Mean Position
When the mass \(M\) passes through the mean position, its velocity is at its maximum (\(v_1\)). At this exact moment, a smaller mass \(m\) is placed on top of it. Since there are no external horizontal forces during this “collision,” momentum is conserved:
\(
M v_1=(M+m) v_2
\)
Where:
\(M=0.9 \mathrm{~kg}=900 \mathrm{~g}\)
\(m=124 \mathrm{~g}\)
\(v_1\) is the initial maximum velocity.
\(v_2\) is the new maximum velocity of the combined system.
Rearranging for the velocity ratio:
\(
\frac{v_1}{v_2}=\frac{M+m}{M}=\frac{900+124}{900}=\frac{1024}{900}
\)
Step 2: Relation Between Velocity and Amplitude
In SHM, the maximum velocity at the mean position is given by \(v_{\text {max }}=\omega A\).
We know \(\omega=\sqrt{\frac{k}{m_{\text {total }}}}\).
For the first case: \(v_1=\sqrt{\frac{k}{M}} A_1\)
For the second case: \(v_2=\sqrt{\frac{k}{M+m}} A_2\)
Taking the ratio \(\frac{v_1}{v_2}\) :
\(
\frac{v_1}{v_2}=\frac{\sqrt{\frac{k}{M}} A_1}{\sqrt{\frac{k}{M+m}} A_2}=\sqrt{\frac{M+m}{M}} \cdot \frac{A_1}{A_2}
\)
Step 3: Combine the Ratios
Equate the results from Step 1 and Step 2:
\(
\begin{aligned}
\frac{M+m}{M} & =\sqrt{\frac{M+m}{M}} \cdot \frac{A_1}{A_2} \\
\frac{A_1}{A_2} & =\sqrt{\frac{M+m}{M}}
\end{aligned}
\)
Substitute the numerical values:
\(
\begin{aligned}
& \frac{A_1}{A_2}=\sqrt{\frac{1024}{900}} \\
& \frac{A_1}{A_2}=\frac{32}{30}=\frac{16}{15}
\end{aligned}
\)
Step 4: Solve for \(\alpha\)
The problem gives the ratio as \(\frac{\alpha}{\alpha-1}\). Comparing this to our result:
\(
\frac{\alpha}{\alpha-1}=\frac{16}{15}
\)
By inspection (or by solving \(15 \alpha=16 \alpha-16\)), we find:
\(
\alpha=16
\)

Hint

(b) To solve this, we need to determine the effective spring constants (\(k_{\text {eff }}\)) for both configurations. In physics problems of this type, the “standard” figures usually show one arrangement in series and the other in parallel.
Step 1: Analyze Figure (a) – Series Arrangement
In a series arrangement, the springs are connected end-to-end. The effective spring constant \(k_s\) is given by:
\(
\begin{gathered}
\frac{1}{k_s}=\frac{1}{k_1}+\frac{1}{k_2}=\frac{1}{k}+\frac{1}{2 k} \\
\frac{1}{k_s}=\frac{2+1}{2 k}=\frac{3}{2 k} \Longrightarrow k_s=\frac{2 k}{3}
\end{gathered}
\)
The time period \(T_a\) is:
\(
T_a=2 \pi \sqrt{\frac{m}{k_s}}=2 \pi \sqrt{\frac{3 m}{2 k}}=3 \mathrm{~s} \dots(1)
\)
Step 2: Analyze Figure (b) – Parallel Arrangement
In a parallel arrangement (or when the mass is between two springs), the springs provide a combined restoring force. The effective spring constant \(k_p\) is:
\(
k_p=k_1+k_2=k+2 k=3 k
\)
The time period \(T_b\) is:
\(
T_b=2 \pi \sqrt{\frac{m}{k_p}}=2 \pi \sqrt{\frac{m}{3 k}} \quad \text { – (Eq. 2) }
\)
Step 3: Find the Ratio and Solve for \(x\)
Divide Eq. 2 by Eq. 1:
\(
\begin{gathered}
\frac{T_b}{T_a}=\frac{2 \pi \sqrt{\frac{m}{3 k}}}{2 \pi \sqrt{\frac{3 m}{2 k}}} \\
\frac{T_b}{3}=\sqrt{\frac{m}{3 k} \cdot \frac{2 k}{3 m}} \\
\frac{T_b}{3}=\sqrt{\frac{2}{9}}=\frac{\sqrt{2}}{3}
\end{gathered}
\)
Solving for \(T_b\) :
\(
T_b=3 \times \frac{\sqrt{2}}{3}=\sqrt{2} \mathrm{~s}
\)
Step 4: Identify the Value of \(x\)
The problem states the period in figure (b) is \(\sqrt{x} \mathrm{~s}\). Comparing this to our result:
\(
x=2
\)

Hint

(d) To find the new amplitude, we need to calculate the velocity of the body at the given position, apply the change, and then use the energy relationship of Simple Harmonic Motion (SHM).
Step 1: Calculate Initial Velocity (\(\boldsymbol{v}_{\mathbf{1}}\))
The velocity of a particle in SHM at a displacement \(y\) is given by:
\(
v=\omega \sqrt{A^2-y^2}
\)
Given the initial amplitude \(A_1=10 \mathrm{~cm}\) and displacement \(y=5 \mathrm{~cm}\) :
\(
\begin{gathered}
v_1=\omega \sqrt{10^2-5^2} \\
v_1=\omega \sqrt{100-25}=\omega \sqrt{75}
\end{gathered}
\)
Step 2: Calculate New Velocity (\(v_2\))
The problem states the velocity was tripled by an air jet:
\(
v_2=3 v_1=3 \omega \sqrt{75}
\)
Step 3: Find the New Amplitude (\(\boldsymbol{A}_{\mathbf{2}}\))
The new velocity \(v_2\) at the same position \(y=5 \mathrm{~cm}\) must satisfy the SHM velocity equation for the new amplitude \(A_2\). Note that \(\omega\) remains the same because the mass and the restoring force constant (the “springiness” of the system) haven’t changed.
\(
v_2=\omega \sqrt{A_2^2-y^2}
\)
Substitute the value of \(v_2\) from Step 2:
\(
3 \omega \sqrt{75}=\omega \sqrt{A_2^2-5^2}
\)
\(
A_2^2=675+25=700
\)
Step 5: Identify the Value of \(x\)
The problem states the new amplitude is \(\sqrt{x} \mathrm{~cm}\). Since \(A_2=\sqrt{700}\), we compare the two:
\(
\begin{gathered}
\sqrt{x}=\sqrt{700} \\
x=700
\end{gathered}
\)

Hint

(a) To solve for the maximum velocity, we use the Law of Conservation of Energy. The potential energy lost by the bob as it falls from its highest point is converted into kinetic energy at the lowest point (the mean position).
Step 1: Find the Vertical Height (\(h\))
When the pendulum is pulled to an angle \(\theta\), it rises a vertical distance \(h\) from its lowest point. Looking at the geometry:
The total length of the string is \(L\).
The vertical distance from the support to the bob’s level at \(60^{\circ}\) is \(L \cos 60^{\circ}\).
The height \(h\) is the difference between the full length and this vertical component.
\(
h=L-L \cos \theta=L(1-\cos \theta)
\)
Given \(L=250 \mathrm{~cm}=2.5 \mathrm{~m}\) and \(\theta=60^{\circ}\) :
\(
\begin{gathered}
h=2.5\left(1-\cos 60^{\circ}\right) \\
h=2.5(1-0.5)=2.5 \times 0.5=1.25 \mathrm{~m}
\end{gathered}
\)
Step 2: Apply Conservation of Energy
At the maximum height, the bob has Potential Energy ( \(P E=m g h\) ) and zero Kinetic Energy. At the lowest point, it has maximum Kinetic Energy ( \(K E=\frac{1}{2} m v^2\) ) and zero Potential Energy relative to that point.
\(
m g h=\frac{1}{2} m v^2
\)
The mass \(m\) cancels out (showing that the velocity is independent of the bob’s weight):
\(
g h=\frac{1}{2} v^2 \Longrightarrow v^2=2 g h
\)
Step 3: Solve for Maximum Velocity (\(v\))
Substitute \(g=10 \mathrm{~m} / \mathrm{s}^2\) and \(h=1.25 \mathrm{~m}\) :
\(
\begin{gathered}
v^2=2 \times 10 \times 1.25 \\
v^2=25 \\
v=\sqrt{25}=5 \mathrm{~ms}^{-1}
\end{gathered}
\)
The maximum velocity attained by the bob will be \(5 \mathrm{~ms}^{-1}\).

Hint

(a) To find the time taken to travel from the maximum displacement to half the amplitude, we use the standard equation of motion for a Simple Harmonic Oscillator.
Step 1: Choose the Equation of Motion
Since the timing starts from the position of maximum displacement (the extreme position), it is most convenient to use the cosine form of the displacement equation:
\(
x=A \cos (\omega t)
\)
Where:
\(x\) is the displacement at time \(t\).
\(A\) is the amplitude (8 cm).
\(\omega\) is the angular frequency.
Step 2: Calculate Angular Frequency (\(\boldsymbol{\omega}\))
The relationship between angular frequency and the time period (\(T=6 \mathrm{~s}\)) is:
\(
\omega=\frac{2 \pi}{T}=\frac{2 \pi}{6}=\frac{\pi}{3} \mathrm{rad} / \mathrm{s}
\)
Step 3: Set up the Displacement Condition
The particle moves from \(x=A\) to \(x=A / 2\). We substitute \(x=A / 2\) into our equation to find the time \(t\) it reaches that point:
\(
\begin{gathered}
\frac{A}{2}=A \cos (\omega t) \\
\frac{1}{2}=\cos (\omega t)
\end{gathered}
\)
Step 4: Solve for Time (\(t\))
We know that \(\cos (\theta)=1 / 2\) when \(\theta=60^{\circ}\) or \(\pi / 3\) radians.
\(
\omega t=\frac{\pi}{3}
\)
Substitute \(\omega=\frac{\pi}{3}\) :
\(
\begin{gathered}
\left(\frac{\pi}{3}\right) t=\frac{\pi}{3} \\
t=1 \mathrm{~s}
\end{gathered}
\)
The time it will take to travel from maximum displacement to half of its amplitude is \(\mathbf{1 s}\).

Hint

(c) To find the value of \(x\), we need to determine the point in time when the energy is split equally between kinetic and potential forms.
Step 1: Set up the Energy Equality
The problem states that at a certain time \(\boldsymbol{t}\), the Kinetic Energy \((K)\) and Potential Energy \((U)\) are equal:
\(
K=U
\)
We also know that the Total Energy (\(E\)) is the sum of both:
\(
E=K+U
\)
Substituting \(K=U\) into the total energy equation:
\(
\begin{gathered}
E=U+U=2 U \\
U=\frac{1}{2} E
\end{gathered}
\)
Step 2: Use the Displacement Relation
The potential energy at any displacement \(y\) is \(U=\frac{1}{2} k y^2\), and the total energy is \(E=\frac{1}{2} k A^2\). Substituting these into our equality:
\(
\begin{aligned}
& \frac{1}{2} k y^2=\frac{1}{2}\left(\frac{1}{2} k \cdot A^2\right) \\
& y^2=\frac{A^2}{2} \Longrightarrow y=\frac{A}{\sqrt{2}}
\end{aligned}
\)
Step 3: Solve for Time (\(t\))
Since the mass was “pulled downward and released,” the motion starts from the extreme position. Therefore, we use the cosine equation for displacement:
\(
y=A \cos (\omega t)
\)
Substitute \(y=\frac{A}{\sqrt{2}}\) :
\(
\begin{aligned}
& \frac{A}{\sqrt{2}}=A \cos (\omega t) \\
& \cos (\omega t)=\frac{1}{\sqrt{2}}
\end{aligned}
\)
We know that \(\cos (\theta)=\frac{1}{\sqrt{2}}\) when \(\theta=\frac{\pi}{4}\).
\(
\omega t=\frac{\pi}{4}
\)
Step 4: Relate to Time Period (\(T\))
Substitute the relation \(\omega=\frac{2 \pi}{T}\) into the equation:
\(
\begin{gathered}
\left(\frac{2 \pi}{T}\right) t=\frac{\pi}{4} \\
t=\frac{\pi}{4} \times \frac{T}{2 \pi} \\
t=\frac{T}{8}
\end{gathered}
\)
Step 5: Identify the value of \(x\)
The problem defines the time as \(t=\frac{T}{x}\). Comparing this to our result:
\(
x=8
\)

Hint

(a) To find the value of \(x\), we need to determine the amplitude of both SHM equations and then calculate their ratio.
Step 1: Find the Amplitude of \(y_1\)
The first equation is given in the standard form \(y=A \sin (\omega t+\phi)\) :
\(
y_1=10 \sin \left(3 \pi t+\frac{\pi}{3}\right)
\)
By direct comparison, the amplitude of the first motion is:
\(
A_1=10
\)
Step 2: Find the Amplitude of \(y_2\)
The second equation is a linear combination of a sine and a cosine function:
\(
y_2=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)
\)
For an expression of the form \(y=a \sin \theta+b \cos \theta\), the resultant amplitude \(R\) is calculated as:
\(
R=\sqrt{a^2+b^2}
\)
In the case of \(y_2\), we have \(a=1\) and \(b=\sqrt{3}\) (inside the parentheses). Let’s calculate the resultant of the terms inside the bracket first:
\(
\text { Bracket Amplitude }=\sqrt{(1)^2+(\sqrt{3})^2}=\sqrt{1+3}=\sqrt{4}=2
\)
Now, multiply by the coefficient 5 outside the bracket to get the total amplitude \(A_2\) :
\(
A_2=5 \times 2=10
\)
Step 3: Calculate the Ratio
We are looking for the ratio of \(A_1\) to \(A_2\) :
\(
\text { Ratio }=\frac{A_1}{A_2}=\frac{10}{10}=1
\)
The problem states the ratio is \(x: 1\). Comparing \(1: 1\) to \(x: 1\), we find: \(x=1\)

Hint

(b) To find the relationship between the two amplitudes, we need to express both equations in the standard SHM form \(x=A \sin (\omega t+\phi)\) and identify their respective amplitudes.
Step 1: Identify the Amplitude of \(x_1\)
The first equation is already in the standard form:
\(
x_1=5 \sin \left(2 \pi t+\frac{\pi}{4}\right)
\)
By inspection, the amplitude of the first motion is:
\(
A_1=5
\)
Step 2: Simplify and Find the Amplitude of \(x_2\)
The second equation is a combination of a sine and a cosine function:
\(
x_2=5 \sqrt{2}(\sin 2 \pi t+\cos 2 \pi t)
\)
For any expression of the form \(y=a \sin \theta+b \cos \theta\), the resultant amplitude \(R\) is given by:
\(
R=\sqrt{a^2+b^2}
\)
Let’s look at the terms inside the parentheses where \(a=1\) and \(b=1\) :
\(
\text { Resultant of }(\sin 2 \pi t+\cos 2 \pi t)=\sqrt{1^2+1^2}=\sqrt{2}
\)
Now, multiply this by the coefficient \(5 \sqrt{2}\) that is outside the parentheses to get the total amplitude \(\boldsymbol{A}_{\mathbf{2}}\) :
\(
\begin{aligned}
& A_2=5 \sqrt{2} \times \sqrt{2} \\
& A_2=5 \times 2=10
\end{aligned}
\)
Step 3: Compare the Amplitudes
We have \(A_1=5\) and \(A_2=10\). To find how many times larger \(A_2\) is compared to \(A_1\) :
\(
\text { Ratio }=\frac{A_2}{A_1}=\frac{10}{5}=2
\)
The amplitude of the second motion is 2 times the amplitude of the first motion.

Hint

(b) To find the value of \(x\), we need to use the initial conditions \((t=0)\) for both the position and velocity functions of the particle.
Step 1: Analyze the Position at \(t=0\)
The displacement function is given as:
\(
x(t)=A \sin (\omega t+\phi)
\)
At \(t=0\), the position is 2 cm :
\(
\begin{gathered}
2=A \sin (0+\phi) \\
2=A \sin \phi-(\text { Eq. } 1)
\end{gathered}
\)
Step 2: Analyze the Velocity at \(t=0\)
The velocity function \(v(t)\) is the derivative of the displacement function:
\(
v(t)=\frac{d x}{d t}=A \omega \cos (\omega t+\phi)
\)
At \(t=0\), the velocity is \(2 \omega \mathrm{~cm} / \mathrm{s}:\)
\(
\begin{gathered}
2 \omega=A \omega \cos (0+\phi) \\
2=A \cos \phi-(\text { Eq. } 2)
\end{gathered}
\)
Step 3: Solve for Amplitude (\(\boldsymbol{A}\))
To eliminate the phase constant \(\phi\), we square both equations and add them together using the identity \(\sin ^2 \phi+\cos ^2 \phi=1\) :
\(
\begin{gathered}
(A \sin \phi)^2+(A \cos \phi)^2=2^2+2^2 \\
A^2\left(\sin ^2 \phi+\cos ^2 \phi\right)=4+4 \\
A^2(1)=8 \\
A=\sqrt{8}=2 \sqrt{2} \mathrm{~cm}
\end{gathered}
\)
Step 4: Identify the Value of \(x\)
The problem states that the amplitude is \(x \sqrt{2} \mathrm{~cm}\). Comparing this to our result:
\(
\begin{gathered}
2 \sqrt{2}=x \sqrt{2} \\
x=2
\end{gathered}
\)
The value of \(x\) is 2.
This result shows that at \(t=0\), the particle was exactly at a phase of \(45^{\circ}(\pi / 4)\), where its potential and kinetic energies are equal. 

Hint

(b) To find the angular frequency (\(\omega\)) of this system, we need to treat it as a two-body oscillator connected by an effective spring system.
Step 1: Determine the Effective Spring Constant (\(k_{\text {eff }}\))
Looking at the setup, the two springs are connected end-to-end between the two masses. This is a series arrangement.
The formula for the effective spring constant (\(k_s\)) in series is:
\(
\frac{1}{k_{e f f}}=\frac{1}{k_1}+\frac{1}{k_2}
\)
Given \(k_1=k=20 \mathrm{~N} / \mathrm{m}\) and \(k_2=4 k=80 \mathrm{~N} / \mathrm{m}\) :
\(
\begin{gathered}
\frac{1}{k_{e f f}}=\frac{1}{20}+\frac{1}{80} \\
\frac{1}{k_{e f f}}=\frac{4+1}{80}=\frac{5}{80}=\frac{1}{16} \\
k_{e f f}=16 \mathrm{~N} / \mathrm{m}
\end{gathered}
\)
Step 2: Calculate the Reduced Mass (\(\mu\))
In a system where two masses are free to move and connected by a spring, the oscillation frequency depends on the reduced mass (\(\mu\)), not just one individual mass.
\(
\mu=\frac{m_A \cdot m_B}{m_A+m_B}
\)
Given \(m_A=200 \mathrm{~g}=0.2 \mathrm{~kg}\) and \(m_B=800 \mathrm{~g}=0.8 \mathrm{~kg}\) :
\(
\begin{gathered}
\mu=\frac{0.2 \times 0.8}{0.2+0.8} \\
\mu=\frac{0.16}{1.0}=0.16 \mathrm{~kg}
\end{gathered}
\)
Step 3: Calculate the Angular Frequency (\(\omega\))
The angular frequency for a two-body oscillator is given by:
\(
\omega=\sqrt{\frac{k_{e f f}}{\mu}}
\)
Substitute the values we calculated:
\(
\begin{aligned}
\omega & =\sqrt{\frac{16}{0.16}} \\
\omega & =\sqrt{100} \\
\omega & =10 \mathrm{rad} / \mathrm{s}
\end{aligned}
\)
The angular frequency of the system is \(\mathbf{1 0 ~ r a d} / \mathbf{s}\).
Key Insight: Using the reduced mass allows us to treat the two-body problem as a single-body problem where one end of the spring is “fixed.” This is why the frequency is higher than if you only considered the larger 800 g mass.

Hint

(b)

To solve for the speed of the bob at \(60^{\circ}\), we use the Law of Conservation of Mechanical Energy. The kinetic energy lost as the bob swings upward is equal to the potential energy it gains.
Step 1: Calculate the Height Gained (\(h\))
When the pendulum string of length \(L\) moves from the vertical to an angle \(\theta\), the vertical height \(h\) it rises is:
\(
h=L-L \cos \theta=L(1-\cos \theta)
\)
Given \(L=50 \mathrm{~cm}=0.5 \mathrm{~m}\) and \(\theta=60^{\circ}\) :
\(
\begin{gathered}
h=0.5\left(1-\cos 60^{\circ}\right) \\
h=0.5(1-0.5)=0.5 \times 0.5=0.25 \mathrm{~m}
\end{gathered}
\)
Step 2: Apply Conservation of Energy
Let \(v_0\) be the speed at the lowest position and \(v\) be the speed at \(60^{\circ}\).
Total Energy at Lowest Point \(=\) Total Energy at \(60^{\circ}\)
\(
\frac{1}{2} m v_0^2=\frac{1}{2} m v^2+m g h
\)
Cancel the mass \(m\) and multiply by 2:
\(
\begin{aligned}
& v_0^2=v^2+2 g h \\
& v^2=v_0^2-2 g h
\end{aligned}
\)
Step 3: Solve for the Final Speed (\(v\))
Substitute the known values \(\left(v_0=3 \mathrm{~m} / \mathrm{s}, g=10 \mathrm{~m} / \mathrm{s}^2, h=0.25 \mathrm{~m}\right)\) :
\(
\begin{gathered}
v^2=(3)^2-2(10)(0.25) \\
v^2=9-5 \\
v^2=4 \\
v=\sqrt{4}=2 \mathrm{~m} / \mathrm{s}
\end{gathered}
\)
The speed of the bob when the length makes an angle \(60^{\circ}\) to the vertical will be \(2 \mathrm{~m} / \mathrm{s}\).

Hint

(d) To find the value of \(a\), we use the displacement equation for a particle starting its motion from the mean position.
Step 1: Choose the Equation of Motion
For a particle starting from the mean position (\(x=0\) at \(t=0\)), the displacement \(x\) at any time \(t\) is given by:
\(
x=A \sin (\omega t)
\)
Where:
\(A\) is the amplitude.
\(\omega\) is the angular frequency.
Step 2: Calculate Angular Frequency (\(\omega\))
The relationship between angular frequency and the time period (\(T=2 \mathrm{~s}\)) is:
\(
\omega=\frac{2 \pi}{T}=\frac{2 \pi}{2}=\pi \mathrm{rad} / \mathrm{s}
\)
Step 3: Set up the Displacement Condition
The particle covers a displacement equal to half of its amplitude (\(x=A / 2\)). We substitute this into our equation:
\(
\begin{gathered}
\frac{A}{2}=A \sin (\omega t) \\
\frac{1}{2}=\sin (\omega t)
\end{gathered}
\)
Step 4: Solve for Time (\(t\))
We know that \(\sin (\theta)=1 / 2\) when \(\theta=30^{\circ}\) or \(\pi / 6\) radians.
\(
\omega t=\frac{\pi}{6}
\)
Substitute \(\omega=\pi\) :
\(
\begin{aligned}
& \pi t=\frac{\pi}{6} \\
& t=\frac{1}{6} \mathrm{~s}
\end{aligned}
\)
Step 5: Identify the Value of \(a\)
The problem states the time taken is \(\frac{1}{a} \mathrm{~s}\). Comparing this to our result:
\(
\begin{aligned}
& \frac{1}{a}=\frac{1}{6} \\
& a=6
\end{aligned}
\)
The value of \(a\) is 6.
Notice that traveling from the mean position to \(A / 2\) takes \(1 / 6\) th of the total time period (\(T / 12\)), whereas traveling from the extreme position to \(A / 2\) takes \(1 / 3\) rd of that time (\(T / 6\)). This is because the particle is moving much faster near the center.

Hint

(b) To solve for the value of \(x\), we need to compare the effective spring constants for a single spring versus two identical springs in a series combination.
Step 1: Analyze Figure 1 (Single Spring)
In the first case, a mass \(M\) is attached to a single spring with spring constant \(k\). The time period \(T_a\) is given by the standard formula:
\(
T_a=2 \pi \sqrt{\frac{M}{k}}
\)
Step 2: Analyze Figure 2 (Series Combination)
In the second case, two identical springs (\(k_1=k\) and \(k_2=k\)) are connected in series. The effective spring constant \(k_s\) for a series combination is:
\(
\begin{gathered}
\frac{1}{k_s}=\frac{1}{k}+\frac{1}{k}=\frac{2}{k} \\
k_s=\frac{k}{2}
\end{gathered}
\)
Now, we calculate the new time period \(T_b\) using this effective constant:
\(
\begin{gathered}
T_b=2 \pi \sqrt{\frac{M}{k_s}}=2 \pi \sqrt{\frac{M}{k / 2}} \\
T_b=2 \pi \sqrt{\frac{2 M}{k}}
\end{gathered}
\)
Step 3: Find the Ratio \(T_b / T_a\)
Now we divide \(T_b\) by \(T_a\) to find the relationship:
\(
\frac{T_b}{T_a}=\frac{2 \pi \sqrt{\frac{2 M}{k}}}{2 \pi \sqrt{\frac{M}{k}}}
\)
The \(2 \pi\) and the \(\sqrt{M / k}\) terms cancel out, leaving:
\(
\frac{T_b}{T_a}=\sqrt{2}
\)
Step 4: Identify the Value of \(x\)
The problem states the ratio is \(T_b / T_a=\sqrt{x}\). Comparing this to our result:
\(
\sqrt{x}=\sqrt{2}
\)
\(
x=2
\)
The value of \(x\) is \(\mathbf{2}\).
By putting springs in series, you essentially made the system “softer” (halved the stiffness), which naturally increased the time it takes for one full oscillation.

Hint

(c) To find the value of \(x\), we need to relate the speed of the particle to its displacement using the standard equations of Simple Harmonic Motion (SHM).
Step 1: Identify Maximum Speed (\(v_{\text {max }}\))
In SHM, the maximum speed of a particle occurs at the mean position and is given by the formula:
\(
v_{\max }=\omega a
\)
Where \(\omega\) is the angular frequency and \(a\) is the amplitude.
Step 2: Use the Velocity-Displacement Relation
The velocity \(v\) of a particle at any displacement \(y\) from the mean position is:
\(
v=\omega \sqrt{a^2-y^2}
\)
Step 3: Set up the Condition
The problem states that the speed \(v\) is half of the maximum speed:
\(
v=\frac{1}{2} v_{\max }
\)
Substituting the expressions from Step 1 and Step 2:
\(
\omega \sqrt{a^2-y^2}=\frac{1}{2}(\omega a)
\)
Step 4: Solve for Displacement (\(y\))
Cancel \(\omega\) from both sides:
\(
\sqrt{a^2-y^2}=\frac{a}{2}
\)
Square both sides to remove the square root:
\(
\begin{gathered}
a^2-y^2=\frac{a^2}{4} \\
y^2=a^2-\frac{a^2}{4} \\
y^2=\frac{3 a^2}{4}
\end{gathered}
\)
Taking the square root of both sides:
\(
y=\frac{\sqrt{3} a}{2}
\)
Step 5: Identify the Value of \(x\)
The problem gives the displacement as \(\frac{\sqrt{x} a}{2}\). Comparing this to our result:
\(
\begin{aligned}
\frac{\sqrt{3} a}{2} & =\frac{\sqrt{x} a}{2} \\
x & =3
\end{aligned}
\)
The value of \(x\) is 3.

Hint

(d) To find the value of \(\alpha\), we need to break down the motion of the pendulum into distinct segments based on the fraction of the oscillation completed.
Step 1: Analyze 5/8 of an Oscillation
One full oscillation \((T)\) consists of four equal segments (quarters):
Mean to Extreme (\(T / 4\))
Extreme back to Mean (\(T / 4\))
Mean to opposite Extreme (\(T / 4\))
Opposite Extreme back to Mean (\(T / 4\))
The fraction \(\frac{5}{8}\) can be written as:
\(
\frac{5}{8}=\frac{4}{8}+\frac{1}{8}=\frac{1}{2}+\frac{1}{8}
\)
This means the pendulum completes:
One half-oscillation: \(\frac{1}{2} T\) (from mean, to one extreme, and back to mean).
An additional \(1 / 8\) oscillation: This is half of a “quarter-swing” (from mean towards the other extreme).
Step 2: Calculate the Time for the Final Segment
The first half-oscillation takes exactly \(\frac{T}{2}\).
Now we need the time \(t\) to cover the remaining distance. In \(1 / 8\) of an oscillation, the particle travels from the mean position to a displacement \(x\). Since a full oscillation is \(4 A\) (where \(A\) is amplitude), \(1 / 8\) of an oscillation is a distance of \(0.5 A\).
Using the displacement equation starting from the mean:
\(
\begin{gathered}
x=A \sin (\omega t) \\
0.5 A=A \sin (\omega t) \Longrightarrow \sin (\omega t)=\frac{1}{2} \\
\omega t=\frac{\pi}{6}
\end{gathered}
\)
Substitute \(\omega=\frac{2 \pi}{T}\) :
\(
\begin{gathered}
\left(\frac{2 \pi}{T}\right) t=\frac{\pi}{6} \\
t=\frac{T}{12}
\end{gathered}
\)
Step 3: Total Time Taken
Sum the time for the half-oscillation and the final segment:
\(
\begin{gathered}
\text { Total Time }=\frac{T}{2}+\frac{T}{12} \\
\text { Total Time }=\frac{6 T+T}{12}=\frac{7}{12} T
\end{gathered}
\)
Step 4: Identify \(\boldsymbol{\alpha}\)
The problem states the time is \(\frac{\alpha}{\beta} T\). Comparing this to \(\frac{7}{12} T\) :
\(\alpha=7\)
\(\beta=12\)