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aleksandr82 [10.1K]
3 years ago
12

A pendulum is set swinging. its first oscillation is through 30 and each succeeding oscillation is through 95% of the angle of t

he one before it. After how many swings is the angle through which it swings less than 1 degree?
Mathematics
1 answer:
Marat540 [252]3 years ago
8 0

<u>Answer-</u>

<em>After 76 swings</em><em> the angle through which it swings less than 1°</em>

<u>Solution-</u>

From the question,

Angle of the first of swing = 30° and then each succeeding oscillation is through 95% of the angle of the one before it.

So the angle of the second swing = (30\times \frac{95}{100})^{\circ}

Then the angle of third swing = (30\times (\frac{95}{100})^2)^{\circ}

So, this follows a Geometric Progression.

(30,\ 30\cdot \frac{95}{100},\ 30\cdot (\frac{95}{100})^2............,0)

a = The initial term = 30

r = Common ratio = \frac{95}{100}

As we have to find the number swings when the angle swept by the pendulum is less than 1°.

So we have the nth number is the series as 1, applying the formula

T_n=ar^{n-1}

Putting the values,

\Rightarrow 1=30(\frac{95}{100})^{n-1}

\Rightarrow \frac{1}{30} =(\frac{95}{100})^{n-1}

Taking logarithm of both sides,

\Rightarrow \log \frac{1}{30} =\log (\frac{95}{100})^{n-1}

\Rightarrow \log \frac{1}{30} =(n-1)\log (\frac{95}{100})

\Rightarrow -1.5=(n-1)(-0.02)

\Rightarrow 1.5=(n-1)(0.02)

\Rightarrow n-1=\dfrac{1.5}{0.02}

\Rightarrow n-1=75

\Rightarrow n=76

Therefore, after 76 swings the angle through which it swings less than 1°

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Solution :

Given :

a = (1, 2, 3, 4) ,    b = ( 4, 3, 2, 1),    c = (1, 1, 1, 1)     ∈   R^4

a). (a.2c)b + ||-3c||a

Now,

(a.2c) = (1, 2, 3, 4). 2 (1, 1, 1, 1)

         = (2 + 4 + 6 + 6)

         = 20

-3c = -3 (1, 1, 1, 1)

     = (-3, -3, -3, -3)

||-3c|| = $\sqrt{(-3)^2 + (-3)^2 + (-3)^2 + (-3)^2 }$

        $=\sqrt{9+9+9+9}$

       $=\sqrt{36}$

        = 6

Therefore,

(a.2c)b + ||-3c||a = (20)(4, 3, 2, 1) + 6(1, 2, 3, 4)  

                          = (80, 60, 40, 20) + (6, 12, 18, 24)

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b). two vectors \vec A and \vec B are parallel to each other if they are scalar multiple of each other.

i.e., \vec A=r \vec B   for the same scalar r.

Given \vec p is parallel to \vec a, for the same scalar r, we have

$\vec p = r (1,2,3,4)$

$\vec p =  (r,2r,3r,4r)$   ......(1)

Let \vec q = (q_1,q_2,q_3,q_4)   ......(2)

Now given \vec p  and  \vec q are perpendicular vectors, that is dot product of \vec p  and  \vec q is zero.

$q_1r + 2q_2r + 3q_3r + 4q_4r = 0$

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Also given the sum of \vec p  and  \vec q is equal to \vec b. So

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We know that,

Position vector = terminal point - initial point

Initial point = terminal point - position point

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                  = (-2, -1, -1, -6)

d). \vec b = (4,3,2,1)

Let us say a vector \vec d = (d_1, d_2,d_3,d_4)  is perpendicular to \vec b.

Then, \vec b.\vec d = 0

     $4d_1+3d_2+2d_3+d_4=0$

     $d_4=-4d_1-3d_2-2d_3$

There are infinitely many vectors which satisfies this condition.

Let us choose arbitrary $d_1=1, d_2=1, d_3=2$

Therefore, $d_4=-4(-1)-3(1)-2(2)$

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