A pendulum clock can be approximated as a simple pendulum of length 1.20m and keeps accurate time at a location where G=9.83m/s2. In a location where g=9.73m/s2, what must be the new length of the pendulum, such that the clock continues to keep accurate time9 that is, its period remains the same)?
Isnt Period on a pendulum clock equal to
2PI sqrt (length/g) ?
so if the period are the same, setting them equal
2PI sqrt (1.2/8.83) = 2PI sqrt (length/9.73)
solve for length.
Yes, you are correct! The period of a simple pendulum can be calculated using the formula you mentioned:
Period = 2π * √(length / g)
In this case, we have the original length of the pendulum, which is 1.20m, and the original value of acceleration due to gravity, which is 9.83m/s^2 in the first location.
Now, we need to find the new length of the pendulum in a location where the acceleration due to gravity is 9.73m/s^2, while keeping the period the same.
By setting the two periods equal to each other, we can solve for the new length. Let's write the equation:
2π * √(1.20 / 9.83) = 2π * √(new length / 9.73)
Now, we can simplify the equation by canceling out the 2π on both sides:
√(1.20 / 9.83) = √(new length / 9.73)
Next, we isolate the new length by squaring both sides of the equation:
1.20 / 9.83 = (new length / 9.73)^2
We can then cross-multiply:
1.20 * (9.73)^2 = (new length)^2 * 9.83
Now, we can solve for the new length by taking the square root of both sides:
new length = √((1.20 * (9.73)^2) / 9.83)
Calculating this expression will give you the value of the new length of the pendulum in the location where the acceleration due to gravity is 9.73m/s^2 while keeping the clock accurate.