If the equation for the displacement of a particle executing simple harmonic motion is x = 3sin(2π/18t + π/6) cm, then the distance travelled by the particle in a time of 36 s is (Time 't' is in second)

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q91 – AP-EAMCET 2025

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Home Physics OscillationsQuestion #380

If the equation for the displacement of a particle executing simple harmonic motion is

x = 3 \sin (\frac{2 π}{18} t + \frac{π}{6})

cm, then the distance travelled by the particle in a time of 36 s is (Time 't' is in second)

Clue 1 of 3: Understanding SHM Parameters

Identify the amplitude and angular frequency from the given displacement equation to determine the time period of oscillation.

🧠 Full Solution Path

Step 1: Identify SHM Parameters✦ Active

The given equation for the displacement of a particle executing simple harmonic motion is $x = 3 \sin (\frac{2 π}{18} t + \frac{π}{6})$ cm. This equation is in the standard form $x = A \sin (ω t + ϕ)$ , where $A$ is the amplitude, $ω$ is the angular frequency, and $ϕ$ is the initial phase.

By comparing the given equation with the standard form, we can identify the amplitude and angular frequency:

A = 3 cm

ω = \frac{2 π}{18} = \frac{π}{9} rad/s

💡 Teacher's Secret Hint

Always ensure the units are consistent. Here, displacement is in cm and time in seconds.

Step 2: Calculate the Time Period○ Expand

The time period ( $T$ ) of simple harmonic motion is related to the angular frequency ( $ω$ ) by the formula:

T = \frac{2 π}{ω}

Substitute the value of $ω$ found in Step 1:

T = \frac{2 π}{\frac{π}{9}} = 2 π \times \frac{9}{π} = 18 s

💡 Teacher's Secret Hint

The time period represents the time taken for one complete oscillation. Make sure to cancel out common terms like $π$ correctly.

Step 3: Calculate Distance in One Period○ Expand

In one complete time period ( $T$ ), a particle executing simple harmonic motion travels a distance equal to four times its amplitude ( $4 A$ ). This is because it moves from its equilibrium position to one extreme ( $A$ ), back to equilibrium ( $A$ ), to the other extreme ( $A$ ), and back to equilibrium ( $A$ ).

Using the amplitude $A = 3$ cm:

Distance in one period = 4 A = 4 \times 3 cm = 12 cm

💡 Teacher's Secret Hint

Remember that displacement is a vector quantity (can be positive or negative), while distance is a scalar quantity and is always positive. For SHM, distance is always $4 A$ for a full cycle.

Step 4: Calculate Total Distance Travelled○ Expand

The total time for which the particle travels is given as $t_{t o t a l} = 36$ s. We need to find out how many full time periods occur within this total time.

Number of periods = \frac{t_{t o t a l}}{T} = \frac{36 s}{18 s} = 2

Since the particle completes 2 full time periods, the total distance travelled will be the number of periods multiplied by the distance travelled in one period.

Total distance = 2 \times (Distance in one period) = 2 \times 12 cm = 24 cm

💡 Teacher's Secret Hint

This method works perfectly for integer multiples of the time period. If the total time were not an integer multiple of $T$ , you would need to consider the phase and position at the start and end of the remaining fraction of a period.

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Physics Question 91 – AP-EAMCET 2025

Clue 1 of 3: Understanding SHM Parameters

🧠 Full Solution Path