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
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.
2 answers
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
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
3 answers