Two simple harmonic motion are represent by the following equations
`y_(1) = 10 sin (3pi t + pi//3) `
`y_(2) = 5 (sin 3 pi t + sqrt3 cos 3 pi t)`
Here `t` is in seconds.
Find out the ratio of their amplitudes.What are the time period of the two motion?

Options

A. `n=(3)/(2)`
B. n=1
C. `m=(2)/(3)`
D. `m=(1)/(2)`

Answer

B

Explanation

(i) The two S.H.M. are given by
`y_(1)=10 "sin" (pi)/(4)(12t+1)`
` =10 "sin"((12 pi t)/(4)+(pi)/(4))`
`y_(1)=10 "sin"(3 pi t+ (pi)/(4))`
`therefore` The amplitude of (1) is 10 units.
and `y_(2)=5 "sin" 3pi t+ 5sqrt(3) "cos" 3pi t`
`therefore " Amplitude " =sqrt(5^(2)+(5sqrt(3))^(2))`
Thus each motion has the amplitude of 10.
`therefore m=(10)/(10) =1`
(ii) In this case, `T=(2pi)/(omega)=(2pi)/(3pi)=(2)/(3)`
and period of (1)= period of (2) `=(2)/(3)s therefore n=1`