At t = 0 a block with mass m = 5 kg moves with a velocity v = 2 m/s at position xo = -.33 m from the equilibrium position of the spring. the block is attached to a massless spring of spring constant k = 61.2 n/m and slides on a frictionless surface. at what time will the block next pass x = 0, the place where the spring is unstretched? t1 =

0·149 s

Explanation:

Initially at t = 0 s, velocity of the block = v = 2 m/s and = - 0·33 m from the equilibrium position of the spring

Spring constant k = 61·2 N/m

Acceleration of the block at any instant of time = - (k × x) ÷ m

Where x is the distance of the block from equilibrium position

A is the amplitude( that is maximum distance from the equilibrium position)

w is the angular frequency

t is the time taken

c is any constant

- w² × x =  - (k × x) ÷ m

From the above equation w = √(k ÷ m) = √(61·2 ÷ 5) = 3·499 rad/s

By substituting the values in the given equation

- 0·33 = A × sin(c) → equation 1

By differentiating the equation x = A × sin(w × t + c) with respect to t on both sides

v = A × w × cos(w × t + c)

By substituting the values

2 = A × w × cos(c) → equation 2

By dividing the equation 1 and equation 2

- (0·33 ÷ 2) × w = tan(c)

tan(c) = - 0·577

⇒ c = π - inverse of tan(0·577)

∴ c = π - 0·523 rad

Substituting the value of c in equation 1

- 0·33 = A × sin(π - 0·523)

∴ A = - 36·67 m

∴ x = - 36·67 × sin(3·499 × t + π - 0·523)

At x = 0

sin(3·499 × t + π - 0·523) = 0

∴ 3·499 × t + π - 0·523 = 0 or π

It can't be 0 because if it is 0, then t is negative

∴ 3·499 × t + π - 0·523 = π

3·499 × t = 0·523

∴ t = 0·523 ÷ 3·499 = 0·149 s

∴ Here = 0·149 s

heats surface water up

it would be c. predictability