Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the o

NCERT Exemplar · Q31

Q. Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator.

Telangana TsbieShort· 2mImportance★★★★★est

86% · 57/66 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Simple Harmonic Motion Energy

Simple Harmonic Motion Energy: From Intuition to Precision

Imagine a pendulum swinging, or a mass bouncing on a spring. You push it once, and it keeps moving back and forth. Where does that energy go? It doesn't vanish — it just changes form. That's the core idea.

The Intuition: A Trade Between Two Forms

Think of a child on a swing. At the highest point, the swing is momentarily still — all the energy is stored as potential energy (the height you could fall from). At the lowest point, the swing is moving fastest — all that stored energy has turned into kinetic energy (the energy of motion). In between, it's a mix of both.

For a spring-mass system (the simplest SHM), the same trade happens:

When the mass is at the extreme position (maximum displacement), it's momentarily at rest — all energy is potential.
When the mass passes through the equilibrium position (the centre), it's moving fastest — all energy is kinetic.
Everywhere else, it's a blend.

The key insight: total mechanical energy stays constant (if no friction). Energy is never created or destroyed — it just shifts between potential and kinetic.

The Precise Statement

For a particle of mass $m$ executing SHM with angular frequency $\omega$ and amplitude $A$ :

E_{\text{total}} = \frac{1}{2} m \omega^2 A^2

This is a constant. At any displacement $x$ from equilibrium:

Kinetic energy: $K = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 (A^2 - x^2)$
Potential energy: $U = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 x^2$ (since $k = m\omega^2$ )
Total energy: $E = K + U = \frac{1}{2} m \omega^2 A^2$

Notice: when $x = \pm A$ , $K = 0$ and $U = E$ . When $x = 0$ , $K = E$ and $U = 0$ .

Why This Matters for Exams

Three things to remember:

Total energy depends only on amplitude and frequency — not on the mass's position or speed at any instant. It's a fixed number for a given oscillation.
Energy is proportional to the square of amplitude: double the amplitude, quadruple the energy. This is a common exam trap — students think doubling amplitude doubles energy. It doesn't.
The potential energy curve is a parabola: $U = \frac{1}{2} k x^2$ . This is why SHM is called "harmonic" — the restoring force ( $F = -kx$ ) comes from this parabolic potential well.

Watch out

A frequent mistake: writing $U = \frac{1}{2} m \omega^2 x^2$ but forgetting that $\omega^2 = k/m$ . Both forms are equivalent — use whichever is given in the problem.

A Quick Check

A mass of 0.5 kg oscillates on a spring with $k = 8$ N/m and amplitude 0.1 m. Find total energy. …

Start your 14-day free trial to unlock the full solution →

The potential energy of a simple harmonic oscillator is half its maximum total energy when its displacement is $\pm \frac{A}{\sqrt{2}}$ , where $A$ is the amplitude.

A simple harmonic oscillator (SHO) is a system that, when displaced from its equilibrium position, experiences a restoring force proportional to the displacement. This leads to oscillatory motion. Key to understanding its energy is recognizing that the total mechanical energy (sum of kinetic and potential energy) remains constant throughout the motion, assuming no damping.

The potential energy (PE) of an SHO is stored due to its displacement from the equilibrium position. It's maximum at the extreme points of oscillation (amplitude $A$ ) and zero at the equilibrium position. The kinetic energy (KE) is maximum at the equilibrium position (where velocity is maximum) and zero at the extreme points (where velocity momentarily becomes zero before reversing direction).

The "maximum energy of the oscillator" refers to this constant total mechanical energy. We can find its value by considering the point where all the energy is potential (at maximum displacement, $x = \pm A$ , where KE is zero) or where all the energy is kinetic (at equilibrium, $x = 0$ , where PE is zero). Since the problem involves potential energy, it's most convenient to use the total energy expressed in terms of the amplitude.

Here's how to find the displacement:

Define the potential energy of a simple harmonic oscillator. The potential energy $PE$ of an SHO at a displacement $x$ from its equilibrium position is given by:

PE = \frac{1}{2}kx^2

where $k$ is the spring constant (or force constant) of the oscillator.

2. Determine the maximum energy of the oscillator.

The total mechanical energy $E$ of an SHO is conserved. At the extreme positions of oscillation, $x = \pm A$ (where $A$ is the amplitude), the oscillator momentarily comes to rest, meaning its kinetic energy is zero. At these points, all the total energy is in the form of potential energy.

Therefore, the maximum energy of the oscillator, which is its total energy, is:

E_{max} = \frac{1}{2}kA^2

This is the constant total energy of the system.

> [!FORMULA]
> The total energy of a simple harmonic oscillator is $E = \frac{1}{2}kA^2$.

3. Set up the condition given in the problem.

The problem states that the potential energy ( $PE$ ) is half of the maximum energy ( $E_{max}$ ).

$PE = \frac{1}{2} E_{max}$ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 20 on this concept.

TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
$2\, \mathrm{Nm\,rad^{-1}}$
›Reveal solutionSolution
The torsional spring constant is found by equating the period formula $T = 2\pi \sqrt{I/\kappa}$ to the given values. Using the moment of inertia of a disc $I = \frac12 m R^2$ , we get $\kappa \approx 1.97\ \mathrm{Nm\,rad^{-1}}$ , so the closest choice is (D).

The key concept is torsional oscillation: a disc suspended by a wire, when twisted and released, undergoes simple harmonic motion. The period depends on the moment of inertia of the disc and the torsional spring constant (the restoring torque per unit twist) of the wire. The formula is exactly analogous to a mass on a spring: $T = 2\pi \sqrt{I/\kappa}$ .

Why this works:

The wire provides a restoring torque $\tau = -\kappa \theta$ , where $\theta$ is the twist angle. Newton’s second law for rotation gives $I \ddot{\theta} = -\kappa \theta$ , which is SHM with angular frequency $\omega = \sqrt{\kappa/I}$ . Hence $T = 2\pi/\omega = 2\pi \sqrt{I/\kappa}$ . So if we know $T$ and $I$ , we can solve for $\kappa$ .

Now, step by step:

Find the moment of inertia of the disc. For a uniform circular disc of mass $m$ and radius $R$ , rotating about an axis through its centre and perpendicular to its plane, the moment of inertia is

$I = \frac12 m R^2.$
Here $m = 10\ \mathrm{kg}$ and $R = 15\ \mathrm{cm} = 0.15\ \mathrm{m}$ .

So
$I = \frac12 \times 10 \times (0.15)^2 = 5 \times 0.0225 = 0.1125\ \mathrm{kg\,m^2}.$

Use the period formula to solve for $\kappa$ . The period is $T = 1.5\ \mathrm{s}$ . From $T = 2\pi \sqrt{I/\kappa}$ , square both sides:

$T^2 = 4\pi^2 \frac{I}{\kappa} \quad\Rightarrow\quad \kappa = \frac{4\pi^2 I}{T^2}.$

Plug in the numbers.

$\kappa = \frac{4\pi^2 \times 0.1125}{(1.5)^2}.$
First compute $4\pi^2 \approx 39.478$ . Then
$\kappa \approx \frac{39.478 \times 0.1125}{2.25} = \frac{4.441275}{2.25} \approx 1.9739\ \mathrm{Nm\,rad^{-1}}.$

Compare with the options. …
Start your 14-day free trial to see the full solution →
TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
$Q. A ladder of length 3 m and mass 20 kg leans on a frictionless wall with its feet at rest on the floor 1 m away from the wall. The reaction force of the wall on the ladder is nearly (A) 34.6 N (B) 98 N (C) 196 N (D) 28.3 N$
›Reveal solutionSolution
The wall is frictionless, so its reaction is purely horizontal. Taking torques about the foot of the ladder gives the wall’s reaction as $R = \frac{mg}{2\tan\theta}$ , which evaluates to about 34.6 N — option (A).

The key idea is that a ladder leaning against a frictionless wall cannot have a vertical reaction from the wall — the wall only pushes horizontally. That means the only vertical forces are the ladder’s weight (downward) and the floor’s normal reaction (upward). The floor also provides friction to keep the foot from slipping, but that friction is not asked for here.

To find the wall’s horizontal push, we use torque equilibrium. The ladder is in static equilibrium, so the net torque about any point must be zero. Choosing the foot of the ladder as the pivot eliminates the torques from both the floor’s normal reaction and the friction force (since they act at the pivot). That leaves only the weight of the ladder and the wall’s reaction to produce torque.

Geometry first. The ladder is 3 m long, and its foot is 1 m from the wall. So the base of the right triangle formed by the ladder, wall, and floor is 1 m, and the hypotenuse is 3 m. The height at which the ladder touches the wall is

$h = \sqrt{3^2 - 1^2} = \sqrt{9 - 1} = \sqrt{8} = 2\sqrt{2} \ \text{m}.$
The angle $\theta$ that the ladder makes with the floor satisfies
$\cos\theta = \frac{1}{3}, \quad \sin\theta = \frac{2\sqrt{2}}{3}, \quad \tan\theta = 2\sqrt{2}.$

Torque about the foot. The weight $mg$ acts at the centre of the ladder, which is 1.5 m from either end. Its lever arm about the foot is the horizontal distance from the foot to the line of action of the weight — that is, half the base: $0.5$ m. The torque due to weight tries to rotate the ladder clockwise (downward).

The wall’s reaction $R$ acts horizontally at the top of the ladder. Its lever arm about the foot is the vertical height $h = 2\sqrt{2}$ m. This torque tries to rotate the ladder counterclockwise.

Equilibrium condition. For zero net torque:
Start your 14-day free trial to see the full solution →
TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
$1 : \sqrt{3}$
›Reveal solutionSolution
The key idea is that in SHM, force is proportional to displacement, and velocity is related to displacement via energy conservation. Given force = 86.6% of maximum, we find displacement = 86.6% of amplitude, then velocity ratio = 1:2, so the correct option is (B).

Concept & Intuition

In simple harmonic motion (SHM), the restoring force is proportional to displacement from equilibrium: $F = -kx$ . The maximum force occurs at the extreme position (amplitude $A$ ), so $F_{\text{max}} = kA$ . When the force is 86.6% of maximum, the displacement is also 86.6% of amplitude. Velocity in SHM is given by $v = \omega \sqrt{A^2 - x^2}$ , and maximum velocity is $v_{\text{max}} = \omega A$ . The ratio $v/v_{\text{max}}$ depends only on $x/A$ . Recognizing 86.6% as $\sqrt{3}/2$ leads directly to the answer.

Step-by-step solution

Relate force to displacement

In SHM, $F = -kx$ . The magnitude of force at displacement $x$ is $|F| = k|x|$ . Maximum force is at $x = \pm A$ , so $F_{\text{max}} = kA$ .

Given $|F| = 86.6\% \text{ of } F_{\text{max}} = 0.866\, kA$ .

Thus $k|x| = 0.866\, kA \Rightarrow |x| = 0.866\, A$ .

Note: $0.866 \approx \sqrt{3}/2$ .

Express velocity in SHM

The velocity at displacement $x$ is $v = \omega \sqrt{A^2 - x^2}$ , where $\omega = \sqrt{k/m}$ .

Maximum velocity occurs at $x = 0$ : $v_{\text{max}} = \omega A$ .

Find the ratio

$\frac{v}{v_{\text{max}}} = \frac{\omega \sqrt{A^2 - x^2}}{\omega A} = \sqrt{1 - \left(\frac{x}{A}\right)^2}.$
Substitute $x/A = 0.866 = \sqrt{3}/2$ :
Start your 14-day free trial to see the full solution →
TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
$Q. A balance is made using a uniform metre scale of mass 100 g and two plates each of mass 200 g fixed at the two ends of the scale and the balance is pivoted at 45 cm mark of the scale. The error when 300 g weight is placed in the plate at 0 cm to weigh vegetables placed in the plate at 100 cm is (A) 36.4 g (B) 63.6 g (C) 200 g (D) 100 g$
›Reveal solutionSolution
Balancing torques about the $45\ \mathrm{cm}$ pivot shows the vegetables actually weigh $200\ \mathrm{g}$ , whereas the $300\ \mathrm{g}$ standard weight is used — an error of $100\ \mathrm{g}$ .

Setting up the balance

Take torques about the pivot at the $45\ \mathrm{cm}$ mark. Masses and their distances from the pivot:

Left plate at $0\ \mathrm{cm}$ : plate $200\ \mathrm{g}$ + weight $300\ \mathrm{g}=500\ \mathrm{g}$ , at $45\ \mathrm{cm}$ to the left.

Right plate at $100\ \mathrm{cm}$ : plate $200\ \mathrm{g}$ + vegetables $W$ , at $55\ \mathrm{cm}$ to the right.

Scale's own weight $100\ \mathrm{g}$ acts at its centre $50\ \mathrm{cm}$ , i.e. $5\ \mathrm{cm}$ to the right of the pivot.

Equilibrium condition

Clockwise (left) torque equals anticlockwise (right) torque:
$500 \times 45 = (200 + W)\times 55 + 100 \times 5$
$22500 = 11000 + 55W + 500$ …
Start your 14-day free trial to see the full solution →
TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
$Q. A balance is made using a uniform metre scale of mass 100 g and two plates each of mass 200 g fixed at the two ends of the scale and the balance is pivoted at 45 cm mark of the scale. The error when 300 g weight is placed in the plate at 0 cm to weigh vegetables placed in the plate at 100 cm is (A) 36.4 g (B) 63.6 g (C) 100 g (D) 200 g$
›Reveal solutionSolution
Taking torques about the pivot at the 45 cm mark, the vegetables that balance a 300 g weight actually have mass 200 g, so the balance over-reads by 100 g. The correct option is (C) 100 g.

Concept

A balance compares torques about its pivot. Because the pivot here is at the 45 cm mark (not at the centre of the system), the two arms are unequal, so a weight placed on one pan does not correspond to an equal mass on the other pan. That mismatch is the "error."

Set-up

Measuring all distances from the pivot at 45 cm:

Uniform scale, mass $100\text{ g}$ , centre of mass at $50\text{ cm}$ → arm $= 50-45 = 5\text{ cm}$ (right of pivot).

Left plate ( $200\text{ g}$ ) at $0\text{ cm}$ → arm $= 45\text{ cm}$ (left).

Right plate ( $200\text{ g}$ ) at $100\text{ cm}$ → arm $= 55\text{ cm}$ (right).

Weight $W = 300\text{ g}$ placed in the $0\text{ cm}$ plate → arm $= 45\text{ cm}$ (left).

Vegetables of mass $m$ in the $100\text{ cm}$ plate → arm $= 55\text{ cm}$ (right). …
Start your 14-day free trial to see the full solution →
TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
$\dfrac{1}{4}$
›Reveal solutionSolution
The disc’s mass equals the extra buoyant force when the immersed fraction increases from 1/4 to 2/5. Using Archimedes’ principle, the mass is 9.6 kg, so option (C) is correct.

Concept & Intuition

Archimedes’ principle tells us that the buoyant force on a floating object equals the weight of the fluid displaced. When the cube floats alone, its weight equals the weight of water displaced by 1/4 of its volume. Adding the disc increases the total weight, so the cube sinks deeper until the new buoyant force (now displacing 2/5 of the cube’s volume) balances the combined weight. The difference in displaced water weight is exactly the weight of the disc.

Step-by-step solution

Find the cube’s volume

Side = 40 cm = 0.4 m

Volume $V = (0.4)^3 = 0.064 \ \text{m}^3$

Buoyant force without the disc

Immersed fraction = $\frac{1}{4}$

Displaced volume $V_1 = \frac{1}{4} \times 0.064 = 0.016 \ \text{m}^3$

Density of water $\rho = 1000 \ \text{kg/m}^3$

Buoyant force = weight of displaced water = $\rho V_1 g$

This equals the cube’s weight $M_{\text{cube}} g$ .

So $M_{\text{cube}} = \rho V_1 = 1000 \times 0.016 = 16 \ \text{kg}$ .

Buoyant force with the disc

New immersed fraction = $\frac{2}{5}$

Displaced volume $V_2 = \frac{2}{5} \times 0.064 = 0.0256 \ \text{m}^3$

Buoyant force = $\rho V_2 g$

This now equals the combined weight $(M_{\text{cube}} + M_{\text{disc}}) g$ .

Find the disc’s mass …
Start your 14-day free trial to see the full solution →
TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
$\dfrac{1}{4}$
›Reveal solutionSolution

Floating equilibrium fixes the cube's mass at $16~\text{kg}$ from the $\tfrac14$ -immersed condition; the extra buoyancy at $\tfrac25$ immersion supports the disc, giving $m=9.6~\text{kg}$ .

Cube of side $L=0.40~\text{m}$ , so volume $V=L^{3}=0.064~\text{m}^{3}$ , with water density $\rho=1000~\text{kg m}^{-3}$ .

Cube alone, $\tfrac14$ immersed — weight balances buoyancy:
$Mg=\rho\left(\frac{V}{4}\right)g\ \Rightarrow\ M=1000\times\frac{0.064}{4}=16~\text{kg}.$
Cube + disc, $\tfrac25$ immersed: …
Start your 14-day free trial to see the full solution →
TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
$50 \ \mathrm{rads}^{-1}$
›Reveal solutionSolution
The angular momentum of a rotating rigid body is $L = I \omega$ . For a solid sphere about its diameter, $I = \frac{2}{5}MR^2$ . Substituting $M = 2\ \mathrm{kg}$ , $R = 0.5\ \mathrm{m}$ , and $\omega = 50\ \mathrm{rad/s}$ gives $L = 10\ \mathrm{Js}$ .

The key idea here is that angular momentum depends on both how the mass is distributed (the moment of inertia) and how fast it spins. For a solid sphere rotating about a diameter, the moment of inertia is a standard result: $I = \frac{2}{5}MR^2$ . This formula comes from integrating the squared distances of all mass elements from the axis of rotation — it’s not something you derive each time in an exam, but you must know it cold.

Once you have $I$ , angular momentum is simply $L = I\omega$ . The units will be $\mathrm{kg\ m^2/s}$ , which is the same as joule-seconds (Js) — so the answer choices are in Js.

Let’s work it through.

Identify the given data

Mass $M = 2\ \mathrm{kg}$

Radius $R = 50\ \mathrm{cm} = 0.5\ \mathrm{m}$ (always convert cm to m for SI units)

Angular speed $\omega = 50\ \mathrm{rad/s}$

Write the moment of inertia for a solid sphere about its diameter

$I = \frac{2}{5} M R^2$

Plug in the numbers

$I = \frac{2}{5} \times 2 \times (0.5)^2 = \frac{2}{5} \times 2 \times 0.25 = \frac{2}{5} \times 0.5 = \frac{1}{5} = 0.2\ \mathrm{kg\ m^2}$

Now compute angular momentum …
Start your 14-day free trial to see the full solution →
TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
$^{-1}$
›Reveal solutionSolution
For a simple harmonic oscillator, the potential energy is $V(x) = \frac12 m \omega^2 x^2$ (plus a constant). Comparing with $V(x) = a + b x^2$ gives $b = \frac12 m \omega^2$ . Substituting $m = 4 \times 10^{-6}$ kg and $\omega = 40$ rad/s yields $b = 3200 \times 10^{-6}$ J/m², so the correct option is (C).

The key idea is that simple harmonic motion arises from a quadratic potential, and the coefficient of $x^2$ is directly related to the mass and angular frequency. The constant term $a$ just shifts the zero of potential energy and doesn't affect the motion.

For a particle of mass $m$ executing SHM with angular frequency $\omega$ , the restoring force is $F = -m\omega^2 x$ , which comes from a potential energy $V(x) = \frac12 m \omega^2 x^2$ (plus any constant). So if we are given $V(x) = a + b x^2$ , then $b$ must equal $\frac12 m \omega^2$ .

Now let's compute it step by step.

Convert mass to SI units. The mass is given as $4$ mg. Since $1$ mg = $10^{-6}$ kg, we have

$m = 4 \times 10^{-6} \ \text{kg}.$

Recall the relation between $b$ and SHM parameters. For SHM, $V(x) = \frac12 m \omega^2 x^2 + \text{constant}$ . Comparing with $V(x) = a + b x^2$ , we identify

$b = \frac12 m \omega^2.$

Plug in the numbers. $\omega = 40$ rad/s, so

$b = \frac12 \times (4 \times 10^{-6}) \times (40)^2.$

Simplify step by step. First, $40^2 = 1600$ . Then …
Start your 14-day free trial to see the full solution →
TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$Q. In a simple pendulum experiment for the determination of acceleration due to gravity, the error in the measurement of the length of the pendulum is 1% and the error in the measurement of the time period is 2%. The error in the estimation of acceleration due to gravity is (A) 1% (B) 3% (C) 4% (D) 5%$
›Reveal solutionSolution
The error in $g$ is the sum of the error in length and twice the error in the period, giving $1\% + 2 \times 2\% = 5\%$ . The correct option is (D).

The key idea is that the acceleration due to gravity $g$ is related to the length $L$ and time period $T$ of a simple pendulum by the formula
$T = 2\pi \sqrt{\frac{L}{g}} \quad \Rightarrow \quad g = \frac{4\pi^2 L}{T^2}.$
When we have percentage errors in measured quantities, we use the rule for propagation of errors in multiplication/division: relative errors add. For a power like $T^2$ , the relative error is multiplied by the exponent. So the error in $g$ is the error in $L$ plus twice the error in $T$ .

Let’s work through it step by step.

Write the formula for $g$ in terms of $L$ and $T$

$g = \frac{4\pi^2 L}{T^2}.$
The constant $4\pi^2$ has no error, so only $L$ and $T$ contribute.

Recall the rule for combining relative errors If $g = k \cdot L^a \cdot T^b$ (with $k$ constant), then the relative error in $g$ is

$\frac{\Delta g}{g} = |a| \frac{\Delta L}{L} + |b| \frac{\Delta T}{T}.$
Here $a = 1$ (for $L$ ) and $b = -2$ (for $T^{-2}$ ), so the absolute values give
$\frac{\Delta g}{g} = 1 \cdot \frac{\Delta L}{L} + 2 \cdot \frac{\Delta T}{T}.$

Plug in the given percentage errors $\frac{\Delta L}{L} = 1\% = 0.01$ and $\frac{\Delta T}{T} = 2\% = 0.02$ . $\frac{\Delta g}{g} = 0.01 + 2 \times 0.02 = 0.01 + 0.04 = 0.05 = 5\%.$ …
Start your 14-day free trial to see the full solution →
TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
$2:3$
›Reveal solutionSolution
In a meter bridge, the balance condition is $R_1/R_2 = L/(100-L)$ . Given $R_1:R_2 = 2:3$ , solving gives $L = 40$ cm. The correct option is (D).

The key idea is that a meter bridge works like a Wheatstone bridge, where the ratio of resistances equals the ratio of the corresponding wire lengths. The wire is 100 cm long, so if the balance point is at $L$ cm from the left, the right segment is $100-L$ cm. The ratio of left gap resistance to right gap resistance is directly the ratio of these lengths.

Why this works:

The meter bridge uses a uniform wire, so resistance is proportional to length. At balance, no current flows through the galvanometer, meaning the potential drop across the left gap resistor equals the drop across the left part of the wire, and similarly for the right side. This gives the simple proportion.

Step-by-step solution:

Set up the balance equation For a meter bridge, if the left gap resistance is $R_1$ and the right gap resistance is $R_2$ , and the balance point is at $L$ cm from the left end, then:

$\frac{R_1}{R_2} = \frac{L}{100 - L}$
This is because the wire is 100 cm long and uniform.

Insert the given ratio The problem states $R_1 : R_2 = 2 : 3$ , so:

$\frac{R_1}{R_2} = \frac{2}{3}$
Therefore:
$\frac{2}{3} = \frac{L}{100 - L}$

Solve for $L$ Cross-multiply: …
Start your 14-day free trial to see the full solution →
TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
$t_1 = (20 \pm 0.4)^\circ \mathrm{C}$
›Reveal solutionSolution
When subtracting measurements, the absolute errors add. The temperature difference is $30°\mathrm{C}$ and the combined error is $\pm 1°\mathrm{C}$ .

Why errors add in subtraction

When you measure two quantities and then subtract them, you might think the uncertainties should also subtract—after all, you're subtracting the values themselves. But that's not how measurement error works.

Each measurement can deviate from its true value in either direction. When you compute the difference, the worst-case scenario occurs when both errors conspire to maximize the deviation. If $t_1$ is measured too high and $t_2$ too low (or vice versa), the errors don't cancel—they reinforce each other. This is why we add the absolute errors even when subtracting measurements.
$Q = A \pm B$
This rule applies to both addition and subtraction of measured quantities.

Step-by-step solution

Identify the measured values and their errors

We have:

$t_1 = 20°\mathrm{C}$ with error $\Delta t_1 = 0.4°\mathrm{C}$

$t_2 = 50°\mathrm{C}$ with error $\Delta t_2 = 0.6°\mathrm{C}$

Calculate the temperature difference

The difference is simply:

$\Delta t = t_2 - t_1 = 50 - 20 = 30°\mathrm{C}$

Find the error in the difference …
Start your 14-day free trial to see the full solution →

←Previous questionQ30 — Show that the motion of a particle represented by y = sinω t - cosω t …→Next questionQ32 — A body of mass m is situated in a potential field U(x) = U₀(1 - cosα x…

Q.Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days