(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?

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

To answer these questions, we'll need to use the appropriate formulas and equations. Let's go through each question step-by-step and explain how to find the answers.

(1) To calculate the percentage decrease in weight, we need to consider the concept of gravitational force. The weight of an object is given by the formula:

Weight = mass * acceleration due to gravity

The acceleration due to gravity can be approximated as:

g = (G * M) / r^2

Where:
G is the gravitational constant (approximately 6.67430 x 10^-11 m^3 kg^-1 s^-2)
M is the mass of the Earth (approximately 5.9722 x 10^24 kg)
r is the distance from the center of the Earth to the object

In this case, the distance from the surface of the Earth to the object is 32 km + 6400 km = 6432 km = 6,432,000 meters.

First, calculate the weight of the body at the Earth's surface using the mass of the body and the acceleration due to gravity at the surface. Then, calculate the weight of the body at a depth of 32 km using the same formula and the new distance from the center of the Earth. Finally, calculate the percentage decrease in weight using the following formula:

Percentage decrease = (Weight at surface - Weight at depth) / Weight at surface * 100

(2) To calculate the frequency of the sound emitted by the wire, we need to use the formula for the fundamental frequency of a stretched string:

Frequency = (1/2L) * sqrt(Tension / (mass per unit length))

Where:
L is the length of the wire
Tension is the tension in the wire
Mass per unit length is the mass of the wire per unit length

In this case, the length of the wire is given as 30 cm = 0.3 m, the tension is 70 N, and the mass per unit length is given as 5.0 g/m = 0.005 kg/m.

Plug these values into the formula to calculate the frequency of the sound emitted by the wire in its fundamental mode.

(3) The interval after which the displacement of the pendulum is half its amplitude can be found by considering the time period of the pendulum. In simple harmonic motion, the time period (T) is given by:

T = 2π * sqrt(l / g)

Where:
l is the length of the pendulum
g is the acceleration due to gravity

In this case, the time period is given as 2 seconds. Rearranging the formula, we can solve for the length of the pendulum:

l = (T / (2π))^2 * g

Once you have the length of the pendulum, you can find the time interval at which the displacement is half its amplitude (A) using the formula:

t = (1/4) * T

Where:
t is the time interval
T is the time period

Substitute the values given in the problem statement into the formulas to calculate the length of the pendulum and the time interval.