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Standard XII

Physics

Acceleration

A block of ma...

Question

A block of mass $2$ kg is attached to the spring of spring constant $50 N m^{- 1}$ . The block is pulled to a distance of $5$ cm from its equilibrium position at $x = 0$ on a horizontal frictionless surface from rest at t = $0$ . The displacement of the block at any time t is then

x = 0.05 s i n 5 t m

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x = 0.05 c o s 5 t m

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x = 0.5 s i n 5 t m

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x = 5 s i n 5 t m

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Solution

The correct option is A $x = 0.05 s i n 5 t m$
Here, $m = 2 k g, k = 50 N m^{- 1}, A = 5 c m = 0.05 m$ . The block exectues SHM. Its angular frequency is given by
$ω = \sqrt{\frac{k}{m}} = \sqrt{\frac{50 N m^{- 1}}{2 k g}} = 5 r a d s^{- 1}$ .
Since the time is noted from the equilibrium position, its displacement at any time t is given by
$x = A s i n ω t = 0.05 s i n 5 t m$

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A block of mass 2kg is attached to the spring of spring constant 50Nm−1. The block is pulled to a distance of 5 cm from its equilibrium position at x=0 on a horizontal frictionless surface from rest at t = 0. The displacement of the block at any time t is then

The correct option is A x=0.05sin5t mHere, m=2kg,k=50Nm−1,A=5cm=0.05m. The block exectues SHM. Its angular frequency is given byω=√km=√50Nm−12kg=5rads−1.Since the time is noted from the equilibrium position, its displacement at any time t is given byx=Asinωt=0.05sin5tm

A block of mass $2$ kg is attached to the spring of spring constant $50 N m^{- 1}$ . The block is pulled to a distance of $5$ cm from its equilibrium position at $x = 0$ on a horizontal frictionless surface from rest at t = $0$ . The displacement of the block at any time t is then