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Determine the value of g if a simple pendulum of length 1 m has a time period of 2 s.


Given: L = 1 m, T = 2 s, g = ?


Description: 68344.png


Description: 68352.png


g = (3.14)2 = 9.8596 m s–2.



A simple pendulum has a time period of 2 s on the earth’s surface. It is taken to a height Re above the earth’s surface, where Re is the radius of the earth. Find the time period at that height.


Let the acceleration due to gravity be g at the earth’s surface and g´ at the said height. At that height, the distance of the pendulum from earth’s centre is Re Re = 2Re.


We know that Description: 68359.png [here Me is the mass of the earth]


Description: 68367.png


Description: 68379.png(1)


The time period of the pendulum at the earth’s surface


Description: 68388.png(2)


The time period at the height Re is


Description: 68398.png(3)


Dividing (3) by (2), we get Description: 68407.png


Description: 68414.png


Description: 68424.png


Therefore, The time period at height Re is 4 s.



Calculate the wavelength of the wave generated by a tuning fork of frequency 256 Hz. The velocity of sound in air at ordinary conditions is 340 m s–1.


Given, f = 256 Hz, v = 340 m s–1


Since, v = f λ


Description: 68435.png



The wavelength and frequency of a sound wave in a certain medium is 40 cm and 825 Hz, respectively. In the same medium, if another wave has the wavelength equal to 32 cm, calculate its frequency.


Given, λ = 40 cm = 0.40 m, f = 825 Hz


v = f  λ = 825 × 0.40 = 330 m s–1.


In the second case, since medium is the same, the velocity of sound will remain the same.


Now, v = 330 m s–1λ = 32 cm = 0.32 m


From v = f λ


Description: 68450.png



A man standing on a terrace hears a thunder. The time interval between a lightning flash and first clap of thunder was found to be 15 s. Calculate the distance of the flash from the observer (speed of sound in air is 332 m s–1).


Since the speed of light is very high, we can assume that the lightning flash is seen at the same instant as it is generated.


Time interval t = 15 s


Speed of sound v = 332 m s–1


Let d be the distance covered by the sound waves. Then,


Description: 68460.png


Description: 68468.png


Therefore, the distance of the observer from the flash is 4980 m.



A stone is dropped into a well 44.1 m deep. The sound of the splash is heard 3.13 s after the stone is dropped. Find the velocity of sound in air. [Take g = 9.8 m s–2]


For vertically downward motion of the stone:


Initial velocity u = 0 (being dropped)


Acceleration due to gravity g = 9.8 m s–2


Distance covered h = 44.1 m


Time taken t = ?


We have


Description: 68478.png


Description: 68501.png ( … u = 0)


Description: 68509.png


Description: 68517.png


Time taken by the sound from the splash to travel from water surface to the edge of the well is 3.13 – 3 = 0.13 s.


For upward motion of the sound waves in air


Description: 68525.png


Description: 68532.png


Therefore, velocity of sound in air is 339.23 m s–1.

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