A pendulum clock has a heavy bob supported on a very thin steel rod that is 1.00000 m long at 18.0celcius.
1 - what is the clock's period if the temperature increases by 11.0 celcius? Assume that coefficient of linear expansion of steel is 1.20×10−5K ^-1 .
2 - The clock keeps perfect time at 18.0celcius. At 29.0celcius after how many hours will the clock be off by 3.00 s?
1) To find the new period of the clock when the temperature increases, we can use the formula:
T2 = T1 * (1 + α * ΔT)
Where:
T2 = new period
T1 = initial period
α = coefficient of linear expansion
ΔT = change in temperature
Given:
T1 = initial period of the clock
α = 1.20×10^(-5) K^(-1) (coefficient of linear expansion of steel)
ΔT = 11.0°C (change in temperature)
We need to find T2, the new period.
First, let's determine the initial period T1.
The period of a pendulum clock can be calculated using the formula:
T1 = 2π * √(L/g)
Where:
T1 = initial period
L = length of the pendulum rod
g = acceleration due to gravity (approximately 9.8 m/s^2)
Given:
L = 1.00000 m (length of the pendulum rod)
g ≈ 9.8 m/s^2
Now, let's calculate T1:
T1 = 2π * √(1.00000 / 9.8)
T1 ≈ 2π * 0.31622776602
T1 ≈ 1.9866375 s
Next, substitute the values into the formula to find T2:
T2 = 1.9866375 * (1 + (1.20×10^(-5) * 11.0))
T2 ≈ 1.9866375 * (1 + 0.000132)
T2 ≈ 1.9866375 * 1.000132
T2 ≈ 1.98777102325 s
Therefore, the clock's new period, when the temperature increases by 11.0°C, is approximately 1.98777102325 seconds.
2) To find the time it takes for the clock to be off by 3.00 seconds when the temperature is 29.0°C, we can use the formula:
Time = (ΔT * T1) / ΔT2
Where:
Time = time in hours
ΔT = change in temperature
T1 = initial period
ΔT2 = change in period
Given:
ΔT = 29.0°C - 18.0°C = 11.0°C (change in temperature)
T1 = initial period of the clock
ΔT2 = 3.00 seconds (change in period)
We need to find Time, the time in hours.
Using the same formula as before, T1 can be calculated:
T1 = 1.9866375 s (as calculated in question 1)
Now, substitute the values into the formula to find the time in hours:
Time = (11.0 * T1) / ΔT2
Time = (11.0 * 1.9866375) / 3.00
Time ≈ 7.3444525 / 3.00
Time ≈ 2.44815083333 hours
Therefore, after approximately 2.44815083333 hours, the clock will be off by 3.00 seconds when the temperature is 29.0°C.
To find the period of the pendulum clock when the temperature increases by 11.0 degrees Celsius, we can use the equation for the period of a pendulum:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
1. First, we need to calculate the new length of the pendulum due to the temperature increase. To do this, we can use the formula for linear expansion:
ΔL = α*L0*ΔT
where ΔL is the change in length, α is the coefficient of linear expansion, L0 is the original length, and ΔT is the change in temperature.
In this case, the original length (L0) is 1.00000 m, the coefficient of linear expansion (α) is 1.20×10^(-5) K^(-1), and the change in temperature (ΔT) is 11.0 degrees Celsius.
ΔL = (1.20×10^(-5) K^(-1)) * (1.00000 m) * (11.0 °C)
ΔL ≈ 1.32×10^(-4) m
The new length of the pendulum (L_new) is the sum of the original length and the change in length:
L_new = L0 + ΔL
L_new = 1.00000 m + 1.32×10^(-4) m
L_new ≈ 1.00013 m
2. Next, we can calculate the new period using the new length:
T_new = 2π√(L_new/g)
To find the new period, we need the acceleration due to gravity (g). Assuming it is approximately 9.8 m/s^2, we can substitute the values into the equation:
T_new = 2π√(1.00013 m / 9.8 m/s^2)
T_new ≈ 2π√0.102 eneb
The new period of the pendulum clock when the temperature increases by 11.0 degrees Celsius is approximately 0.6376 seconds.
Now let's move on to the second question:
To calculate the time it takes for the clock to be off by 3.00 seconds when the temperature is 29.0 degrees Celsius, we need to find the change in period first.
1. Using the same method as above, we find the new length of the pendulum when the temperature is 29.0 degrees Celsius:
ΔL = (1.20×10^(-5) K^(-1)) * (1.00000 m) * (29.0 °C)
ΔL ≈ 3.48×10^(-4) m
L_new = 1.00000 m + 3.48×10^(-4) m
L_new ≈ 1.00035 m
2. Next, we calculate the new period:
T_new = 2π√(L_new/g)
T_new = 2π√(1.00035 m / 9.8 m/s^2)
T_new ≈ 2π√0.1021 seconds
3. Now, we can determine how many seconds the clock gains or loses per period:
ΔT = T_new - T
ΔT ≈ 2π√0.1021 s - 2π√0.102 s
ΔT ≈ 0.0147 s
The clock gains 0.0147 seconds per period. To find out how many hours it takes for the clock to be off by 3.00 seconds, we can set up a proportion:
0.0147 s / 1 period = 3.00 s / x hours
Cross-multiplying, we get:
0.0147 s * x hours = 1 period * 3.00 s
Simplifying:
x ≈ (1 period * 3.00 s) / 0.0147 s
Using the value of the period we found earlier, we can calculate x:
x ≈ (0.6376 s * 3.00 s) / 0.0147 s
x ≈ 130.20 hours
Therefore, the clock will be off by 3.00 seconds after approximately 130.20 hours.