A motor cycle starts from rest and accelerates along a straight path at 2 ms^-2. At the starting poi

Q: Why do we use a minus sign in the numerator of the Doppler formula?

Because the observer is moving away from the source. This relative motion decreases the rate at which the sound wave crests hit the observer, leading to a lower perceived frequency. Hence, we subtract the observer's velocity from the speed of sound.

Q: Does the acceleration of the motorcycle affect the Doppler formula?

No, the Doppler effect depends only on the instantaneous velocity at the exact moment the sound is heard. The acceleration is only used later in the kinematics part to find how far the motorcycle traveled to reach that velocity.

Q: Why did we cancel $f$ from both sides? Don't we need the actual frequency?

The problem gives the apparent frequency as a percentage ($94\%$) of the original frequency. Because we are working with a ratio ($f'/f = 0.94$), the actual value of $f$ doesn't matter and mathematically cancels out.

Q: Can we use $s = ut + \frac{1}{2}at^2$ instead to find the distance?

You could, but you would first need to find the time $t$ using $v = u + at$. Using the third equation of motion ($v^2 = u^2 + 2as$) is a more direct, one-step approach since we don't know (and don't need to know) the time.

Q: What if the wind was blowing? Would that change the answer?

Yes! Wind velocity would add to or subtract from the speed of sound $v$ relative to the ground, modifying the effective speed of sound in the Doppler formula. However, in standard textbook problems like this, we assume still air unless stated otherwise.

Question

A motor cycle starts from rest and accelerates along a straight path at $2 	ext{ ms}^{-2}$. At the starting point of the motor cycle, there is a stationary electric siren. How far has the motor cycle gone when the driver hears the frequency of the siren at 94% of its value when the motor cycle was at rest? (Speed of sound $= 330 	ext{ ms}^{-1}$)

Options:

(A) 49 m

(B) 98 m

(C) 147 m

(D) 196 m

Sigma Percentile · Accepted Answer

### The Doppler Effect Meets Kinematics: A Motorcycle's Journey

Imagine you are sitting on a motorcycle, right next to a blaring electric siren. The siren is stationary, but you rev the engine and accelerate away from it at a constant rate of $2 	ext{ m/s}^2$. As you speed up, you notice something interesting: the pitch of the siren sounds lower and lower. This is the classic Doppler effect in action!

Our goal is to find out exactly how far you have traveled when the perceived frequency drops to $94\%$ of its original value. This problem is a beautiful blend of wave physics and classical mechanics.

### The Physics of Sound

To connect the perceived frequency with the motorcycle's speed, we use the Doppler effect formula. Since the source (the siren) is stationary ($v_s = 0$) and the observer (you on the motorcycle) is moving away, the apparent frequency $f'$ is given by:

$$f' = f \left( \frac{v - v_o}{v} ight)$$

Here, $v$ is the speed of sound ($330 	ext{ m/s}$), and $v_o$ is the instantaneous speed of the observer. We are told that the apparent frequency is $94\%$ of the original frequency, which means $f' = 0.94 f$. Let's substitute these values into our equation:

$$0.94 f = f \left( \frac{330 - v_o}{330} ight)$$

### Finding the Speed

Notice that the original frequency $f$ appears on both sides of the equation. We can simply cancel it out, which tells us that the actual pitch of the siren doesn't matter—only the ratio of the frequencies does. This leaves us with:

$$0.94 = \frac{330 - v_o}{330}$$

Now, let's separate the terms on the right side. Dividing $330$ by $330$ gives $1$, so we have:

$$0.94 = 1 - \frac{v_o}{330}$$

Rearranging this to solve for the velocity term, we get:

$$\frac{v_o}{330} = 1 - 0.94 = 0.06$$

Multiplying $0.06$ by $330$ gives us the velocity of the motorcycle at that exact moment:

$$v_o = 0.06 	imes 330 = 19.8 	ext{ m/s}$$

### The Kinematics Connection

Now that we know the final velocity ($v_o = 19.8 	ext{ m/s}$), and we know the motorcycle started from rest ($u = 0$) with an acceleration of $a = 2 	ext{ m/s}^2$, we can find the distance traveled. We use the third equation of motion:

$$v_o^2 = u^2 + 2as$$

Substituting our known values into the kinematics equation:

$$(19.8)^2 = 0^2 + 2(2)s$$

Squaring $19.8$ gives us $392.04$. On the right side, $2 	imes 2$ is $4$. So, we have:

$$392.04 = 4s$$

Dividing by $4$, we find that the distance $s$ is exactly:

$$s = \frac{392.04}{4} = 98.01 	ext{ m}$$

Rounding off to the nearest integer, the motorcycle has traveled approximately **$98 	ext{ m}$**. This perfectly matches option (b). It is incredibly satisfying to see how seamlessly the physics of waves and the physics of motion come together to solve a real-world problem!

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Doppler Effect Formula$

$Substituting Values$

$Simplifying the Equation$

$Solving for v_{o}$

$Calculating v_{o}$

$Kinematics Equation$

$Substituting into Kinematics$

$Calculating Distance s$

$Final Result$

$The Way Forward$

The Sigma Insight: Doppler Effect

The Doppler Effect Meets Kinematics

The Physics of Sound

Finding the Speed

The Kinematics Connection

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Doppler Effect Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for vo​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating vo​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kinematics Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting into Kinematics

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Distance s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Result

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward