A 0.810 m long brass pendulum experiences a 18.0°C temperature increase.
(a) Calculate the original period(s).
(b) What is the change in the period(s)?
(c) Find the ratio of the change in period to the original period, and calculate how long (h) a clock run by this pendulum would take to fall 1.00 s behind the correct time.
T₀=2πsqrt(L/g) =
=2πsqrt(0.81/9.8)=
=1.8064 s.
ΔL=α•L•ΔT=19•10⁻⁶•0.81•18
=2.77•10⁻⁴m
L₁=L+ ΔL =0.81+2.77•10⁻⁴=
=0.810277
T₁=2πsqrt(L₁/g)=
=2πsqrt(0.810277/9.8)=
=1.8067 s
T₁-T=1.8067-1.8064=0.0003 s.
(T₁-T)/T=0.0003/1.8064=1.66•10⁻⁴
(t/T₁)0.0003= 1 s
t= T₁/0.0003 = 1.8067/0.0003 =6022 s=1.67 h
L1 = 0.810 m.
Temp. Change = +18o C.
T.C. = 1.9*10^-5/C
a. T^2 = 4pi^2(L/g)
T^2 = 39.4(0.81/9.8) = 3.25653
T = 1.8046 s.
b. L2 = L1 + L1*(T.C.)*18
L2=0.81 + 0.81(1.9*10^-5)*18=0.81027702
m.
T^2 = 39.4(0.81027702/9.8)=3.25764
T = 1.8049 s.
Change in period = 1.8049-1.8046 = 0.00003 s.
c. 3*10^-5/1.8046 = 1.6624*10^-5
To answer these questions, we will use the formula for the period of a simple pendulum:
T = 2π√(L/g)
where:
T = period of the pendulum,
L = length of the pendulum, and
g = acceleration due to gravity.
(a) To calculate the original period (T₀), we will use the given length (L = 0.810 m) and the average value for the acceleration due to gravity (g = 9.81 m/s²).
Substituting the values into the formula, we get:
T₀ = 2π√(L/g)
T₀ = 2π√(0.810/9.81)
T₀ ≈ 2.860 s
Therefore, the original period is approximately 2.860 seconds.
(b) To find the change in the period (ΔT), we need to consider the effect of temperature on the length of the pendulum. The change in period is given by the equation:
ΔT = β × T₀ × ΔTᵀ
where:
β = temperature coefficient of linear expansion for brass,
ΔTᵀ = change in temperature, and
T₀ = original period.
The temperature coefficient of linear expansion for brass (β) is typically around 19 × 10^(-6) °C^(-1). The change in temperature (ΔTᵀ) is given as 18.0°C.
Substituting the values into the equation, we get:
ΔT = (19 × 10^(-6) °C^(-1)) × (2.860 s) × (18.0°C)
ΔT ≈ 0.00919 s
Therefore, the change in period is approximately 0.00919 seconds.
(c) The ratio of the change in period to the original period can be calculated as:
Ratio = ΔT / T₀
Ratio = 0.00919 s / 2.860 s
Ratio ≈ 0.00322
To calculate how long the clock will take to fall 1.00 second behind the correct time, we can use the equation:
h = ΔTᵀ × Ratio
Substituting the values into the equation, we get:
h = (1.00 s) × (0.00322)
h ≈ 0.00322 s
Therefore, the clock would take approximately 0.00322 seconds or 3.22 milliseconds to fall 1.00 second behind the correct time.
To calculate the original period of the pendulum, we can use the formula for the period of a simple 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.
(a) For part (a), we need to find the original period of the pendulum. The information given states that the length of the pendulum is 0.810 m. We can assume that the acceleration due to gravity is approximately 9.8 m/s².
T₀ = 2π√(L/g)
T₀ = 2π√(0.810/9.8)
T₀ ≈ 2 × 3.1416 × √(0.08265)
T₀ ≈ 2.0284 s
So, the original period of the pendulum is approximately 2.0284 seconds.
(b) To find the change in the period, we need to know how temperature affects the length of the pendulum. The formula for the change in length due to temperature is:
ΔL = αL₀ΔT
where ΔL is the change in length, α is the coefficient of linear expansion for the material, L₀ is the original length, and ΔT is the change in temperature.
To find the change in period, we need to know how the change in length affects the period. For small displacements, the period of a simple pendulum is approximately given by:
T = 2π√(L/g)
Taking the derivative of this equation with respect to L, we get:
dT/dL ≈ -π/(gL)
So, the change in period due to a small change in length is approximately given by:
ΔT = -T₀(dT/dL)ΔL
Substituting the values:
ΔT = -T₀(-π/(gT₀))ΔL
ΔT = πiven that the length of the pendulum is 0.810 m and the temperature increases by 18.0°C, we can find the change in length:
ΔL = αL₀ΔT
ΔL = (19 × 10^(-6) K^(-1)) × (0.810 m) × (18.0°C)
ΔL ≈ 2.6818 × 10^(-4) m
Now, we can calculate the change in period:
ΔT = πΔL/g
ΔT = π × (2.6818 × 10^(-4)) / 9.8
ΔT ≈ 2.7455 × 10^(-5) s
So, the change in period is approximately 2.7455 × 10^(-5) seconds.
(c) To find the ratio of the change in period to the original period, we can use the formula:
Ratio = ΔT / T₀
Ratio = (2.7455 × 10^(-5) s) / (2.0284 s)
Ratio ≈ 1.3540 × 10^(-5)
Therefore, the ratio of the change in period to the original period is approximately 1.3540 × 10^(-5).
To calculate how long a clock run by this pendulum would take to fall 1.00 s behind the correct time, we can use the formula:
Δt = T₀ - (T₀ + ΔT)
where Δt is the time difference, T₀ is the original period, and ΔT is the change in period.
Δt = 2.0284 s - (2.0284 s + 2.7455 × 10^(-5) s)
Δt ≈ -2.7454 × 10^(-5) s
Since the time difference is negative, we can say that the clock would be 1.00 s ahead of the correct time, not behind.
Therefore, a clock run by this pendulum would be 1.00 second ahead of the correct time.