(1) body is taken to depth of 32km below the surface of the earth.
Calculate the percentage decrease in the weight of the body at this depth
(The radius of the earth is 6400km)
(2)A wire of mass per unit length 5.0gm-1 is stretched between 2 points 30cm apart.
The tension in the wire is 70N.
Calculate the frequency of the sound emitted by the wire when it oscillates in the fundamental mode.
(3)The periodic time of the pendulum executing simple harmonic motion is 2s. After how much interval from t = 0, will it displacement behalf of it amplitude?
Physics - drwls, Saturday, July 26, 2008 at 5:12pm
1) The weight of the person is proportional to the mass of the Earth at lower values of the radius. This will be less by a factor of (6368/6400)^3 at the 32 km depth.
2). Use the wide lineal density and the tension to compute the transverse wave speed, V.
V = sqrt [Tension/(mass per length)]
The length of the wire is half the wavelength of the fundamental mode. Call that wavelength W.
W * f = the wave speed, V. Solve for f
3) displacement = (Amplitude) sin (2 pi t/P)
where P is the period. If displacement= amplitude/2, then
1/2 = sin (2 pi t/P)
2 pi t/P = pi/6 (30 degrees)
t/P = 1/12
t = 1/6 second