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Question
A 100 gm block is connected to a horizontal massless spring of force constant 25.6 N/m and is free to oscillate on a horizontal frictionless surface. The block is displaced on a horizontal frictionless surface. The block is displaced by 3 cm along positive x - direction and released with an initial velocity of vi=−16√3 cm.
Express position of block as function of time.
A 100 gm block is connected to a horizontal massless spring of force constant 25.6 N/m and is free to oscillate on a horizontal frictionless surface. The block is displaced on a horizontal frictionless surface. The block is displaced by 3 cm along positive x - direction and released with an initial velocity of vi=−16√3 cm.
Express position of block as function of time.
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Solution
The correct option is B
So far tile is at x=3 cm
And has v=−16√3 cm/s
Hmmmm....... So where will the particle be if its seen as a projection of a particle doing uniform
Circular motion.
Since the particle is coming back it will be at B
Keeping this in mind Let's start.
Given m=0.1 kg
K=25.6 N/m
We know k=mω2⇒ω=16
Particle is at x=3
v=−16√3
From the above diagram we know ψ is greater then 90∘
So far tile is at x=3 cm
And has v=−16√3 cm/s
Hmmmm....... So where will the particle be if its seen as a projection of a particle doing uniform
Circular motion.
Since the particle is coming back it will be at B
Keeping this in mind Let's start.
Given m=0.1 kg
K=25.6 N/m
We know k=mω2⇒ω=16
Particle is at x=3
v=−16√3
From the above diagram we know ψ is greater then 90∘