If x - a cos nt - b sin nt - then d2x-dt2-

The correct option is A n^2x x=acos( nt ) nbsp; bsin( nt ) nbsp; nbsp;⇒ dx / dt = nasin( nt ) nb cos( nt ) nbsp; nbsp;⇒ d ^ 2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Higher Order Derivatives

If x = a co...

Question

If $x = a cos n t - b sin n t$ , then $\frac{d^{2} x}{d t^{2}} =$

n^{2} x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- n^{2} x

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- n x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $- n^{2} x$
$x = a cos (n t) - b sin (n t)$
$\Rightarrow \frac{d x}{d t} = - n a sin (n t) - n b cos (n t)$
$\Rightarrow \frac{d^{2} x}{{d t}^{2}} = - n^{2} a (cos (n t)) + n^{2} b cos (n t)$
$= - n^{2} (a cos n t - b sin n t) = - n^{2} x$

Suggest Corrections

Similar questions

Q. If x = a cos nt − b sin nt, then

\frac{d^{2} x}{d t^{2}}

is

(a) n² x
(b) −n² x
(c) −nx
(d) nx

Q. Let

f (x) lim n \to \infty {⎛ ⎜ ⎜ ⎝ \frac{n^{n} (x + n) (x + \frac{n}{2}) . . . (x + \frac{n}{n})}{n! (x^{2} + n^{2}) (x^{2} + \frac{n^{2}}{4}) . . . (x^{2} + \frac{n^{2}}{n^{2}})} ⎞ ⎟ ⎟ ⎠}^{\frac{x}{n}}

, for all x>0. Then

Let f (x) = lim n \to \infty {⎛ ⎜ ⎜ ⎝ \frac{n^{n} (x + n) (x + \frac{n}{2}) \dots (x + \frac{n}{n})}{n! (x^{2} + n^{2}) (x^{2} + \frac{n^{2}}{4}) \dots (x^{2} + \frac{n^{2}}{n^{2}})} ⎞ ⎟ ⎟ ⎠}^{\frac{x}{n}},

for all

x > 0.

Then

When n resistances each of value R are connected in parallel, then resultant resistance is x. When these n resistances are connected in series, total resistance is

List - I	List - II
a) $A r {+ N}_{2} X^{-} C u B r - -- \to H B r$	1)ArF
b) $A r {+ N}_{2} X^{-} H B F_{4} - -- \to Δ$	2)ArH
c) $A r {+ N}_{2} X^{-} C_{2} H_{5} O H - ---- \to$	3)ArOH
d) $A r {+ N}_{2} X^{-} H_{2} O - - \to$	4)ArBr

Join BYJU'S Learning Program

If x=acosnt−bsinnt , then d2xdt2=

The correct option is A −n2xx=acos(nt)−bsin(nt)⇒dxdt=−nasin(nt)−nbcos(nt)⇒d2xdt2=−n2a(cos(nt))+n2bcos(nt)=−n2(acosnt−bsinnt)=−n2x

If $x = a cos n t - b sin n t$ , then $\frac{d^{2} x}{d t^{2}} =$

The correct option is A $- n^{2} x$
$x = a cos (n t) - b sin (n t)$
$\Rightarrow \frac{d x}{d t} = - n a sin (n t) - n b cos (n t)$
$\Rightarrow \frac{d^{2} x}{{d t}^{2}} = - n^{2} a (cos (n t)) + n^{2} b cos (n t)$
$= - n^{2} (a cos n t - b sin n t) = - n^{2} x$