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प्रश्न
The motion of a particle of mass m is given by x = 0 for t < 0 s, x(t) = A sin 4 pt for 0 < t < (1/4) s (A > o), and x = 0 for t > (1/4) s. Which of the following statements is true?
- The force at t = (1/8) s on the particle is – 16π2 Am.
- The particle is acted upon by on impulse of magnitude 4π2 A m at t = 0 s and t = (1/4) s.
- The particle is not acted upon by any force.
- The particle is not acted upon by a constant force.
- There is no impulse acting on the particle.
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उत्तर
a, b and d
Explanation:
Given, `x = 0` for `t < 0 s`.
`x(t) = A sin 4pit`; for `0 < t < 1/4 s`
`x = 0`; for `t > 1/4 s`
For, `0 < t < 1/4 s`
`v(t) = (dx)/(dt) = 4π A cos 4πt`
`a(t)` = acceleration
= `(dv(t))/(dt)`
= – 16π2 A sin 4 πt
At `t = 1/8 s, a(t) = - 16 π^2 A sin 4π xx 1/8 = - 16π^2 A`
`F = ma (t) = - 16 π^2` A × m = `- 16 π^2` mA
Impulse = Change in linear momentum
= `F xx t = (- 16 π^2 Am) xx 1/4`
= `- 4π^2 Am`
The impulse (Change in linear momentum) at t = 0 is the same as, t = `1/4 s`.
Clearly, force depends upon A which is not constant. Hence, force is also not constant.
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