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Sagot :
a) 120 vibrations per minute
b)d) 2 Hz (vibrations per second)
c) The period is half of the second.
b)d) 2 Hz (vibrations per second)
c) The period is half of the second.
Explanation :
It is given that an oscillator makes 360 oscillations in 3 minutes.
(a) Using unitary method :
No. of vibrations in one minute is, [tex]n=\dfrac{360}{3}=120[/tex]
So, no of vibrations in one minute is 120.
(b) Similarly,
3 minutes = 180 seconds
No of vibrations in one second is [tex]\dfrac{360}{180}=2[/tex]
So, the no of vibrations in one second is 2.
(c) Time period of the wave is given by :
[tex]T=\dfrac{1}{2\ s^{-1}}[/tex]
[tex]T=0.5\ s[/tex]
The time period of the wave is 0.5 s
(d) The no of vibation per second is called as its frequency.
[tex]\nu=\dfrac{1}{T}[/tex]
[tex]\nu=2\ Hz[/tex]
Hence, this is the required solution.
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