Question

The differential equation of s.h.m of mass 2 g is given by d²x/dt²+16x=0, find the force constant, period and frequency of oscillation?

Answer 1

Answer:

Given:

Equation of SHM has been given as :

$\dfrac{ {d}^{2}x }{d {t}^{2} } + 16x = 0$

To find:

• Force constant
• Time Period
• Frequency

Concept:

First let's try to derive the equation .

For any SHM , we can say :

$\bigstar \:\:F = - kx \:$

$= > F + kx = 0$

Dividing by mass on both sides :

$= > \dfrac{F}{m} + \dfrac{kx}{m} = 0$

$= > a + \dfrac{kx}{m} = 0$

In terms of differentiation :

$= > \dfrac{ {d}^{2}x }{d {t}^{2} } + \dfrac{kx}{m} = 0$

Comparing with the given Equation , we get :

$\boxed{ \dfrac{k}{m} = 16}$

Let time period be T

$T = 2\pi \sqrt{ \dfrac{m}{k} }$

$= > T = 2\pi \sqrt{ \dfrac{1}{16} }$

$= > T = 2\pi (\dfrac{1}{4} )$

$= > T = \dfrac{\pi}{2} \: sec$

So final answer is :

$\boxed{\green{\bold{T = \dfrac{\pi}{2} \: sec}}}$

Frequency is reciprocal of Time Period :

$\boxed{\blue{\bold{freq. = \dfrac{1}{T} = \dfrac{2}{\pi} \: hz}}}$

Now ,

force constant :

$\dfrac{k}{m} = 16$

$= > \dfrac{k}{2 \times {10}^{ - 3} } = 16$

$= > k = 32 \times {10}^{ - 3} N{m}^{-1}$

So final answer :

$\boxed{\red{\bold{\large{k = 32 \times {10}^{ - 3} N{m}^{-1}}}}}$

Answer 2

Answer:

• The Force Constant (K) is 0.032 N/m
• Time period (T) of the oscillator is π/2 Sec.
• Frequency (f) of the oscillator is 2/π Hz.

Given:

1. Mass of the particle = 2 g = 2 × 10⁻³ Kg
2. Differential equation ; d² x / d t² + 16 x = 0

Explanation:

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From the given differential equation,

⇒ d² x / d t² + 16 x = 0

⇒ d² x / d t² = - 16 x

We know that, d²x/dt² = a

⇒ a = - 16 x

⇒ - ω² x = - 16 x  ∵ [ a = - ω² x ]

⇒ ω² = 16

⇒ ω = √16

⇒ ω = 4 rad/sec __[1]

From the formula we know,

⇒ ω² = K / m

⇒ K = ω² × m

Substituting the values,

⇒ K = (4)² × 2 × 10⁻³

⇒ K = 16 × 0.002

⇒ K = 0.032

⇒ K = 0.032 N/m

∴ The Force Constant (K) is 0.032 N/m.

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From the equation (1)

⇒ ω = 4 rad/sec.

⇒ ω = 4

⇒ 2 π / T = 4  ∵ [ ω = 2 π / T ]

⇒ 4 × T = 2 π

⇒ T = 2 π / 4

⇒ T = π / 2

⇒ T = π / 2 Sec.

∴ Time period (T) of the oscillator is π/2 sec.

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From the relation we know,

⇒ f = 1 / T

Substituting the values,

⇒ f = 1 / (π/2)

⇒ f = 1 × 2 / π

⇒ f = 2 / π

⇒ f = 2 / π Hz.

∴ Frequency (f) of the oscillator is 2/π Hz.

Note:

• Symbols have their usual meanings.

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