Question

Body executing shm along a straight line has a velocity of 3 metre per second when it is at a distance of 4 metre per second from its mean position and 4 metre per second when it is at a distance of 3 m from its mean position its angular frequency in aptitude

Answer 1

There are two possibilities to this :

1. V= Wsqrt(A^2 - X^2)

Put V= 4,X=3 (get equation 1) V=3, X= 4 (get equation 2)

Solve 2 equation in 2 variables get A= 5, W=1.

2. V=w√[a^2-x^2]

A simple harmonic motion when the acceleration is directly proportional to displacement in negative direction and a=-w^2y

Answer 2

Answer:

mean position=5m

w = 1 rad

Explanation:

v^2=w^2(a^2-x^2)

then put value,

(1).... v=3,x=4

9 = w^2(a^2-16)

w^2•a^2 = 16w^2 + 9 ..... (1)

(2)..... v=4,x=3

16 = w^2(a^2-9)

w^2•a^2 = 9w^2 +16 .....(2)

eqn(1)&eqn(2)

16w^2+9 =9w^2+16

7w^2 = 7

w^2 = 1

w=1

and mean position

X1=4,X2=3

mean position = √ (x1)^2 + (x2)^2

mean position = √16+9

mean position. = √25

mean position = 5

hence,

w = 1 rad

mean position= 5m