Question

A block of mass 2 kg is attached to a massless spring of spring constant 50 N/m. The block is
pulled to a distance x= 10 cm from its equilibrium position at x=0, on a smooth horizontal
surface from rest at t=0. What is the expression for its velocity ?
a) 5 cos (5t)
b) 5 sin (5t) c) 0.5 cos (5t) d) 2 x 10-2 cos (10t)​

Answer 1

Question :-

A block of mass 2 kg is attached to a massless spring of spring constant 50 N/m. The block is pulled to a distance x= 10 cm from its equilibrium position at x=0, on a smooth horizontal surface from rest at t=0. What is the expression for its velocity ?

Answer :-

Option (c) is correct.

$\implies \: \red{ \boxed{\bf{v \: = 0.5 \: \cos(5t) }}}$

Step - by - step explanation :-

Formula used :-

Here we used equation of simple harmonic motion.

$\implies \: \boxed{ \bf{ y \: = r \: \sin( \omega \: \: t) }} \\ \\ \star \: \bf{ y \: is \: displacement }\: \: \\ \\ \star \: \bf{r \: is \: amplitude} \\ \\ \star \: \bf{ \omega \: is \:angular \: frequency }$

And ,

$\boxed{ \bf{velocity \: = \frac{dy}{dt} }}$

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Solution :-

Given that,

• Mass (m) = 2 kg
• Spring constant (k) = 50 N/m

We know that,

$\boxed{ \bf{ Angular \: frequency \: ( \omega) = \sqrt{ \frac{k}{m} } }} \\ \\ \implies \: \omega \: = \sqrt{ \frac{50}{2} } \\ \\ \implies \: \omega \: = \sqrt{25} \\ \\ \implies \: \omega \: = 5$

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Now ,using equation of S.H.M.

$\implies \: \bf{ \: y \: = 10 \sin(5t) }$

Therefore,

$\implies \: \bf{ velocity \: (v) \: = \frac{dy}{dt} } \\ \\ \implies \: \bf{ v \: = \frac{d}{dt} \big(10 \sin(5t) \big) } \\ \\ \implies \: \bf{ v \: = 10 \cos(5t) \frac{d}{dt} 5t} \\ \\ \implies \: \bf{v \: = 50 \: \cos(5t) } \:$

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In metre,

$\implies \: \bf{v \: = 50 \times {10}^{ - 2} \: \: \cos(5t) } \\ \\ \implies \: \: \boxed{\bf{v \: = 0.5 \cos(5t) }}$

This is the required solution.

Answer 2

The correct option is (a) "5 cos (5t)".

Explanation:

The general equation of simple harmonic motion is given by :

$y=A\sin\omega t$

y is displacement

A is amplitude of wave

$\omega$ is angular frequency

Angular frequency is :

$\omega=\sqrt{\dfrac{k}{m}}$

Spring constant is 50 N/m

Mass of the block is 2 kg

$\omega=\sqrt{\dfrac{50}{2}}\\\\\omega=5$

The block is  pulled to a distance x= 10 cm from its equilibrium position at x=0, on a smooth horizontal  surface from rest.

Equation of SHM becomes :

$y=10\sin(5t)$

Velocity of the block in SHM is given by :

$v=\dfrac{dy}{dt}\\\\v=\dfrac{d(10\sin(5t))}{dt}\\\\v=10\ \cos(5t)\times 5\\\\v=50\ \cos(5t)$

So, the correct option is (a) "5 cos (5t)".

Learn more,

Simple harmonic motion

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