The function ft satisfies the differential equationd2fdt2-f-0 and the auxiliary conditions- f 0 -0- dfdt0-4- The Laplace transform of ft is given by

Given differential equation is d^2fdt^2+f=0...(1) Also f(0)=0,dfdt(0)=4 Taking laplace transform of equation (1), we get ⇒ L[d^2fdt^2+f]=0 ⇒ s^2F(s) s.0 4+F(s)= ...

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Standard XII

Physics

Young's Modulus of Elasticity

The function ...

Question

The function f(t) satisfies the differential equation $\frac{d^{2} f}{d t^{2}} + f = 0$ and the auxiliary conditions, $f (0) = 0, \frac{d f}{d t} (0) = 4$ . The Laplace transform of f(t) is given by

\frac{2}{s + 1}

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\frac{4}{s + 1}

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\frac{4}{s^{2} + 1}

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\frac{2}{s^{4} + 1}

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Solution

The correct option is C $\frac{4}{s^{2} + 1}$
Given differential equation is
$\frac{d^{2} f}{d t^{2}} + f = 0... (1)$
Also $f (0) = 0, \frac{d f}{d t} (0) = 4$
Taking laplace transform of equation (1), we get
$\Rightarrow L [\frac{d^{2} f}{d t^{2}} + f] = 0$
$\Rightarrow s^{2} F (s) - s .0 - 4 + F (s) = 0$
$\Rightarrow F (s) = \frac{4}{s^{2} + 1}$ , the laplace transform of f(t).

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The function f(t) satisfies the differential equationd2fdt2+f=0 and the auxiliary conditions, f(0)=0,dfdt(0)=4. The Laplace transform of f(t) is given by

The correct option is C 4s2+1Given differential equation is d2fdt2+f=0...(1) Also f(0)=0,dfdt(0)=4 Taking laplace transform of equation (1), we get ⇒L[d2fdt2+f]=0 ⇒s2F(s)−s.0−4+F(s)=0 ⇒F(s)=4s2+1, the laplace transform of f(t).

The function f(t) satisfies the differential equation $\frac{d^{2} f}{d t^{2}} + f = 0$ and the auxiliary conditions, $f (0) = 0, \frac{d f}{d t} (0) = 4$ . The Laplace transform of f(t) is given by