if a pendulum clock keeping at sea level is taken at depth h=1km below sea level then the clock approximately gives time is a.gain 13.5 s per day b.loses 13.5s per day c.loses 7 s per day d.gains 7 s per day
T = 2 pi sqrt(L/g)
I could compute T for the two values of g but the change in g is small so better to use calculus
for sport let's make L = 1 meter
T = 2 pi sqrt L (g^-.5)
=2 pi sqrt(1/9.81)
= 2 Seconds for L = 1 and g = 9.81
dT/dg = 2 pi sqrt L(-.5)(g^-1.5)
so dT = - pi dg /g^1.5
Now what is dg ?
g = 9.81 (Re^2/r^2) where Re is R earth
dg/dr = 9.81 Re^2 (-2 r)/r^4
so
dg = 9.81 (Re^2/r^2)(-2/r) dr
for this small change in r, r is about Re and dr = -1000 meters
so
dg = 9.81 (-2/6.4*10^6) (-10^3)
dg = 3.06*10^-3
now use that dg to find the change in our period
Remember we got:
dT = - pi dg /g^1.5
so
dT = - pi (3.06*10^-3) /(9.81)^1.5
dT= - .313 * 10^-3
the negative sign means the pendulum swings faster , shorter period
our fractional change in period is
-.313*10^-3 seconds/ 2 seconds
=-.156*10^-3 seconds per second
multiply that by the number of seconds in a day to find out how much it gets ahead in a day
-.156*10^-3 * 3600 * 24 = 13.5 seconds per day fast
Answers : (1)
At sea level,
gravitational constant that is
g = GM/R
and Time period,
T = 2pi*root(l/g)
at below 1 km,
g’ = GM/(R-1000)
T’ = 2pi*root(l/g’)
T/T’ = root(g’/g) = > root(R/(R-1000))
T’ = T/ root(R/(R-1)) => T/root(6400/6399) = > T/root(1.000156274) => T/1.000078134
or
T’ = 0.999921872*T
putting value of T for appropriaate T we get
T’ approx 13.6 seconds.
When a pendulum clock is taken to a depth below sea level, the gravitational acceleration acting on the clock will be different than at sea level.
The acceleration due to gravity decreases with depth below the Earth's surface. This is because the mass of the Earth is concentrated towards the center, resulting in a decrease in gravitational acceleration as you move further away from the surface.
The change in gravitational acceleration affects the motion of the pendulum, which in turn affects the timekeeping of the clock.
According to the formula 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 gravitational acceleration.
As the gravitational acceleration decreases with depth, the period of the pendulum will increase, causing the clock to lose time.
Therefore, the correct answer is b. The clock loses 13.5 seconds per day when taken 1 km below sea level.
To determine how the pendulum clock at sea level behaves when taken to a depth of 1 km below sea level, we need to consider the effect of gravity.
Gravity affects the period of a pendulum, which in turn affects the time measured by the clock. The formula for the period of a pendulum is:
T = 2π√(L/g)
Where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
At sea level, the acceleration due to gravity is approximately 9.8 m/s². When the pendulum clock is taken to a depth of 1 km below sea level, the value of g decreases because the gravitational field strength decreases with depth.
To calculate the value of g at a depth of 1 km below sea level, we can use the following formula:
g' = g - (change in g with depth)
The change in g with depth can be calculated using the formula:
(change in g with depth) = (gravitational field gradient) * (depth)
The gravitational field gradient is approximately equal to 10 m/s² per km. Substituting the values into the formula, we get:
(change in g with depth) = 10 m/s² per km * (1 km) = 10 m/s²
Now, let's calculate the new value of g:
g' = 9.8 m/s² - 10 m/s² = -0.2 m/s²
Since the new value of g is negative, it means the pendulum clock takes longer to complete a swing, resulting in a longer period and a slower passage of time.
Since the period of the pendulum clock increases, it loses time. To determine the exact amount of time lost, we can use the formula:
(loss of time per day) = (change in period) * (number of periods in a day)
The change in period can be calculated using the formula:
(change in period) = (new period) - (initial period),
where the initial period can be assumed to be approximately 2 seconds (because most pendulum clocks have a period around 2 seconds).
The new period can be calculated using the formula for the period of a pendulum mentioned earlier. Substituting the values into the formula, we get:
T' = 2π√(L/g') = 2π√(L/(-0.2))
Now, we can plug in these values and calculate the loss of time per day:
(loss of time per day) = (2π√(L/(-0.2))) - 2) * (number of periods in a day)
Assuming there are approximately 86,400 seconds in a day and dividing this by the initial period of 2 seconds, we get the number of periods in a day:
(number of periods in a day) = 86,400 s / 2 s ≈ 43,200
Thus, we have:
(loss of time per day) ≈ ((2π√(L/(-0.2))) - 2) * 43,200
Calculating this expression will give us the approximate loss of time per day for the pendulum clock taken to a depth of 1 km below sea level.
Therefore, the correct answer cannot be determined without knowing the length of the pendulum (L).