Question

A body executing SHM has a velocity of 7m/s at a distance of 4m from the mean position and 5m/s at a distance of 5m from the mean position . Find the amplitude and time period of SHM​

Answer 1

Answer:

Explanation:

Equation of the velocity,

v²=ω²(a²−y²)

where,

v = velocity,

ω = angular frequency

a =  amplitude  

y = displacement from the mean position  

=> Here, A body executing SHM has a velocity of 7m/s at a distance of 4m from the mean position

7² =  ω² (a²- 4²)

49=ω²(a²−16) ...(1)

=> A body executing SHM has a velocity of 5m/s at a distance of 5m from the mean position.

5²=ω²(a²−5²)      

25=ω²(a²−25) ...(2)

=> dividing equation (2) by equation (1), we get,

25/49 = a² - 25 / a² - 16

25 (a² - 16) = 49 (a² - 25)

25a² - 400 = 49a² - 1,225

1,225 - 400 = 49a² - 25a²

825 = 24a²

a² = 34.375

a =  5.86

=> now putting the value of a in eqn (1), we get

49=ω²(34.375−16)

49 = ω² * 18.375

ω² = 49 / 18.375

ω² = 2.666

ω = 1.63 rad/s .  

=> Time period of SHM​:

T = 2π / ω

= 2 * 3.14 / 1.63

= 3.85 s

Thus, the amplitude and time period of SHM​ is 1.63 rad/s and 3.85 s.