i will begreatfull if u help me insolving the below problems:
1.)the time period of a pendulum clock on the surface of the earth is 6 sec.Find the timeperiod of the same pendulum on the moon.(g=9.8m/sec2)
2.)the time period of a pendulum is 6.28sec.Whay will be its lengthif g=9.8m/sec2
3.)a long spring whose one end is rigidly fixed is stretched from the other end and then left.longitudinal waves of frequency 10Hz are produced .if the velocity of the wave is 40m/sec ,find the distance between two consecutive compressions in the spring.
1) Period is inversely proportional to g. On the moon, g is 9.8/6 = 1.63 m/s^2.
Pendulum period is 6s*sqrt6 = 14.7 s.
2) Solve P = 6.28 s = 6.28 sqrt(L/g)
L/g = 1 s^2
L = 9.8 m
3) Wavelength = (Wave speed)/(frequency)
= (40 m/s)/10 s^1) = 4 m
T =6 s. g = 9.8 m/s², T2=?
1.
Acceleration due to gravity at the Moon is
g1 = G•M1R1²,
where the gravitational constant G =6.67•10^-11 N•m²/kg²,
Mass of the moon M1= 7.36•10^22 kg,
Radius of the Moon R1 = 1.74 •10^6 m.
T =2•π•sqrt(L/g),
T1 = 2•π•sqrt(L/g1).
T/T1 =sqrt(g1/g),
T1 = T/sqrt(g1/g),
2. T=6.28 s. L=?
T =2•π•sqrt(L/g),
L = (T/2•π)² •g
3.
λ = v/f =40/10 = 4 m
x = λ/2 = 2 m
Sure, I'd be happy to help you solve these problems. Let's break them down one by one:
1) To find the time period of the pendulum on the moon, we need to use the formula for the time period of a pendulum, which is given by:
Time Period = 2π√(L/g)
where L is the length of the pendulum and g is the acceleration due to gravity.
Since we are given the time period on Earth (6 seconds), the length (L), and the value of g (9.8 m/s^2), we can rearrange the formula to solve for L:
L = (Time Period)^2 * g / (4π^2)
Substituting the values, we get:
L = (6^2) * 9.8 / (4π^2)
L = 352.8 / (4π^2)
L ≈ 8.92 meters
So, the time period of the same pendulum on the moon would be approximately 8.92 seconds.
2) In this problem, we have the time period of the pendulum (6.28 seconds) and the value of g (9.8 m/s^2). We need to find the length of the pendulum using the formula mentioned earlier.
Rearranging the formula, we have:
L = (Time Period)^2 * g / (4π^2)
Substituting the given values, we get:
L = (6.28^2) * 9.8 / (4π^2)
L ≈ 1.00 meters
So, the length of the pendulum would be approximately 1.00 meter.
3) To find the distance between two consecutive compressions in the spring, we need to use the formula for the velocity of a wave, which is given by:
Velocity = Frequency * Wavelength
In this problem, we are given the frequency (10 Hz) and the velocity (40 m/s). We need to solve for the wavelength.
Rearranging the formula, we have:
Wavelength = Velocity / Frequency
Substituting the given values, we get:
Wavelength = 40 / 10
Wavelength = 4 meters
So, the distance between two consecutive compressions in the spring would be 4 meters.
I hope this helps! Let me know if you have any further questions.