Log functions:
A new car has an interior sound level of 70 dB at 50 km h. A second car, at the
same speed, has an interior sound level that is two times more intense than that
of the new car. Calculate the sound level inside the second car.
NEW CAR:
10LogN = 70db,
Divide boh sides by 10:
LogN = 7,
N = 10^7.
2ND CAR:
N = 2*10^7,
10Log(2*10^7) = 10*(0.3010+7) = 73db.
Well, well, well, looks like we have some sound level shenanigans going on here! Let's clown around with some logarithms to solve this mystery.
First, we know that the interior sound level of the new car is 70 dB. Now, we need to find the sound level inside the second car.
To do that, we can use the fact that sound intensity is logarithmically related to decibels. In other words, we can use the formula:
dB = 10 * log(I/I₀),
where dB is the sound level in decibels, I is the sound intensity, and I₀ is a reference intensity level.
Since the second car has a sound level that is two times more intense than the new car, we can say that I₂ = 2 * I₁, where I₂ is the sound intensity in the second car and I₁ is the sound intensity in the new car.
Now, let's solve for the sound level in the second car using logarithm magic!
dB₂ = 10 * log(I₂/I₀)
= 10 * log(2 * I₁/I₀)
= 10 * [log(2) + log(I₁/I₀)]
= 10 * [log(2) + dB₁/10]
Where dB₂ is the sound level in the second car and dB₁ is the sound level in the new car (which we already know is 70 dB).
Plugging in the values:
dB₂ = 10 * [log(2) + 70/10]
= 10 * [log(2) + 7]
≈ 10 * [0.3010 + 7]
≈ 10 * 7.301
≈ 73.01 dB.
So, the sound level inside the second car is approximately 73.01 dB. Keep those clown ears protected!
To solve this problem, we can use the fact that the sound level is measured in decibels (dB) and is proportional to the logarithm of the sound intensity.
Step 1: Understand the problem
We are given that the interior sound level of the new car is 70 dB at 50 km/h. We need to find the sound level inside the second car, which has an intensity that is two times more intense than the new car.
Step 2: Use the formula for sound level in decibels
The formula for sound level in decibels is given by:
L = 10 * log10(I/I0)
Where L is the sound level in decibels, I is the sound intensity, and I0 is the reference intensity.
Step 3: Identify the reference point
The reference point for sound level measurements is typically the threshold of hearing, which is approximately 10^(-12) W/m^2. This means that the reference intensity (I0) is 10^(-12) W/m^2.
Step 4: Calculate the sound intensity in the new car
Given that the sound level in the new car is 70 dB, we can substitute the values into the formula to find the sound intensity (I) in the new car:
70 = 10 * log10(I/10^(-12))
Step 5: Solve for the sound intensity in the new car
Dividing both sides of the equation by 10 and taking the inverse log on both sides, we have:
log10(I/10^(-12)) = 7
Taking the inverse log of both sides gives:
I/10^(-12) = 10^7
Simplifying the equation, we get:
I = 10^(-12 + 7)
I = 10^(-5) W/m^2
Step 6: Calculate the sound intensity in the second car
Since the second car has an intensity that is two times more intense than the new car, we need to multiply the sound intensity in the new car by 2.
I2 = 2 * I
Plugging in the value of I that we found in step 5, we get:
I2 = 2 * 10^(-5)
I2 = 2 * 10^(-5) W/m^2
Step 7: Calculate the sound level in the second car
Using the formula for sound level in decibels (L = 10 * log10(I/I0)), we can calculate the sound level in the second car:
L2 = 10 * log10(I2/I0)
Plugging in the values we found in step 6 and the reference intensity (I0 = 10^(-12)), we have:
L2 = 10 * log10((2 * 10^(-5))/10^(-12))
Simplifying the expression inside the logarithm, we get:
L2 = 10 * log10(2 * 10^7)
Using the logarithmic property log(a * b) = log(a) + log(b), we can simplify further:
L2 = 10 * (log10(2) + log10(10^7))
Since log10(10^7) = 7, we have:
L2 = 10 * (log10(2) + 7)
Calculating the value inside the parentheses, we get:
L2 = 10 * (0.3010 + 7)
Simplifying the expression inside the parentheses further, we have:
L2 = 10 * 7.3010
L2 ≈ 73 dB
Therefore, the sound level inside the second car is approximately 73 dB.
To calculate the sound level inside the second car, we can use the concept of logarithms. Specifically, we can use the logarithmic relationship between sound levels and intensity.
The relationship between sound level (measured in decibels, dB) and intensity (measured in watts per square meter, W/m^2) is given by the formula:
dB = 10 * log(I / I0)
where dB is the sound level, I is the intensity being measured, and I0 is a reference intensity level (usually 10^(-12) W/m^2, which corresponds to the threshold of human hearing).
In this case, we know that the first car has a sound level of 70 dB at 50 km/h. Let's denote the intensity of the first car as I1. We want to find the sound level of the second car, which has an intensity (let's call it I2) that is two times more intense than that of the first car.
1. Calculate the intensity of the first car (I1):
Using the formula, we can rearrange it to solve for intensity:
I1 = I0 * 10^(dB / 10)
Using the given information, we have:
dB1 = 70
Using the reference intensity I0 = 10^(-12) W/m^2, we can calculate the intensity of the first car:
I1 = (10^(-12)) * 10^(70 / 10)
2. Calculate the intensity of the second car (I2):
We are told that the intensity of the second car is two times more intense than that of the first car. So we can write:
I2 = 2 * I1
3. Calculate the sound level of the second car (dB2):
Using the formula, we can find the sound level of the second car:
dB2 = 10 * log(I2 / I0)
Substituting the calculated values of I2 and I0 in the formula, we can find the sound level of the second car.
Note: It's important to keep in mind that this calculation assumes that the sound is being measured under the same conditions (e.g. same distance from the source of the sound) in both cars.