## What is the wavelength of a sound wave with frequency of 50hz?

Answer: The wavelength of a sound wave is 6.84 meters.

## What is the frequency of a wave that has a wavelength of 2 m and a speed of 2 m s?

Calculating Wave Frequency or Wavelength from Wave Speed For example, suppose that a wave is traveling at a speed of 2 meters per second and has a wavelength of 1 meter. Then the frequency of the wave is: Frequency = 2m/s1m=2 waves/s, or 2 Hz. Q: A wave is traveling at a speed of 2 m/s and has a frequency of 2 Hz.

## How do you calculate simple pendulum?

A mass m suspended by a wire of length L is a simple pendulum and undergoes simple harmonic motion for amplitudes less than about 15º. The period of a simple pendulum is T=2π√Lg T = 2 π L g , where L is the length of the string and g is the acceleration due to gravity.

## What is the length of simple pendulum which takes 2 seconds?

The seconds pendulum (also called the Royal pendulum), 0.994 m (39.1 in) long, in which each swing takes one second, became widely used in quality clocks. The long narrow clocks built around these pendulums, first made by William Clement around 1680, became known as grandfather clocks.

## Can a simple pendulum be used in an artificial satellite?

Answer. A simple pendulum cannot be used in an artificial satellite because: In an artificial satellite acceleration due to gravity is zero so the pendulum will not vibrate inside it. The pendulum moves due to gravity.

## What is the length of seconds pendulum on moon?

Therefore, the time period of a second’s pendulum on the surface of the moon will be 2√6s.

## What is the period of a pendulum on the moon?

Problem 2 – A pendulum clock on the moon has a length of 2 meters, and its period is carefully measured to be 7.00 seconds. What is the acceleration of gravity on the moon? Answer: 7.00 = 2 π (2/g)1/2 so g = 8π2/49 so g = 1.61 meters/sec2. So the difference in period will be 0.000014 seconds or 14 microseconds.

## What would be the length of a second pendulum at a planet?

∴l′=l4. This means that the required length of the seconds pendulum on the other planet is equal to l4.