Question

"The motion of 10 g mass tied to massless spring is
represented by S.H.M. x = 25 cos 3t+pi/4
where
is in cm and t in second, the force constant of the sprin
is :
a) 9 Nm-
(b) 0.9 Nm-
C) 0.09 Nm-
(d) none of the above.​

Answer 1

Answer:

• The Force Constant (K) of the spring is 0.09 N / m.

Given:

1. The Given Equation is x = 25 cos ( 3 t + π /4 )
2. Mass of the spring (m) = 10 grams.

Explanation:

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From standard equation we know,

$\dashrightarrow \mathrm{x = A\sin(\omega t + \phi)}$

By comparing it with the given equation.

We get,

$\dashrightarrow \mathrm{A = 25 \; cm}\\\\\dashrightarrow \rm{\omega = 3 \; rad/sec}\\\\\dashrightarrow \rm{\phi = \dfrac{\pi}{4}}$

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From the relation we know,

$\dashrightarrow\mathrm{K = \omega^2\; M}$

Where,

• K is Force Constant.
• ω is Angular velocity.
• M is mass of body.

Substituting,

$\dashrightarrow\mathrm{K = \omega^2\; M} \\\\\\ \dashrightarrow\mathrm{K = (3)^2\times 10 \; grams}\\\\\\ \dashrightarrow\mathrm{K = (3)^2\times \dfrac{10}{1000} \ Kg \ \ \because 10 grams = \dfrac{10}{1000} \ Kg}\\\\\\\dashrightarrow\mathrm{K = 9\times \dfrac{1}{100}}\\\\\\\dashrightarrow{\underline{\boxed{\red{\rm K = 0.09 \; N/m}}}}$

∴ The Force Constant (K) of the spring is 0.09 N / m.

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Answer 2

$\Huge{\mathbb{SOLUTION \colon }}$

• Spring Constant Of The Spring is 0.09 N/m

Given,

• Mass Of The Spring,M = 10g

• The position of the particle is given as :

$\tt \: x = 25 \: \cos(3t \: + \dfrac{\pi}{4} ) - - - - - - - - - (1)$

To finD

• Spring Constant (K)

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General Equation Of SHM is

$\tt \: x = a \: \cos( \omega \: t \: + \phi) - - - - - - - (2)$

Comparing equations (1) and (2),we get :

• Amplitude,A = 25 cm

• Angular Velocity = 3 rad/s

• Phase Angle = π/4

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Using the Relation,

$\huge{\boxed{\boxed{\tt K = \omega^2 M }}}$

Substituting the values,we get :

$\longrightarrow \tt K = (3)^2 \times 10^{-2} \\ \\ \longrightarrow \tt K = 9 \times 10^{-2} \\ \\ \large{\longrightarrow \boxed{\boxed{\tt K = 0.09 \ N/m }}}$

Spring Constant is 0.09 N

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