Question

1)The ratio of the kinetic energy and potential energy possessed by a body executing SHM when it is a distance of 1 th/n of it's amplitude from mean position is (potential energy at mean position is assumed to be zero)
1) n²
2)1/n²
3) n² +1
4) n²-1

2)A body is projected vertically up with certain velocity and reaches the ground after 12 sec The ratio of displacement covered by the body in 2 nd second and 7 th second is

1)9:1
2)1:3
3)1:5
4)data not sufficient
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Answer 1

ANSWER 1:

• Option 4) n² - 1.

GIVEN:

• The body is executing Simple Harmonic Motion.

• It is at a distance of 1 / n th of it's amplitude from mean position.

TO FIND:

• The ratio of the kinetic energy and potential energy possessed by the body.

EXPLANATION:

Let the amplitude be A and the distance be D.

$\boxed{\large{\bold{Potential \ Energy = \dfrac{1}{2}m\omega^2(A^2 - D^2)}}}$

$\boxed{\large{\bold{Kinetic\ Energy = \dfrac{1}{2}m\omega^2 \left( \frac{1}{ {D}^2 }\right)}}}$

$\sf\dfrac{Kinetic\ energy }{Potential\ energy}=\dfrac {\dfrac{1}{2}m\omega^2 \left( \dfrac{1}{ {D}^2 }\right)}{\dfrac{1}{2}m\omega^2(A^2-D^2)}$

$\sf \dfrac{Kinetic \ energy }{Potential\ energy}=\dfrac{\left( \dfrac{1}{ {D}^2}\right)} {(A^2-D^2)}$

$\sf \dfrac{Kinetic \ energy }{Potential\ energy}=\dfrac{A^2-D^2}{ {D}^2 }$

$\sf \dfrac{Kinetic \ energy }{Potential\ energy}=\dfrac{A^2}{ {D}^2 } - \dfrac{D^2}{ {D}^2}$

$\sf \dfrac{Kinetic \ energy }{Potential\ energy}= \dfrac{A^2}{ {D}^2 } - \dfrac{D^2}{ {D}^2}$

$\sf \dfrac{Kinetic \ energy }{Potential\ energy}= \left( \dfrac{A}{ {D}}\right)^{2} -1$

$\sf\large{D = \dfrac{A}{n}}$

$\sf\large{n =\dfrac{A}{D}}$

$\bf{Substituting \ n = \dfrac{A}{D}}$

$\sf \dfrac{Kinetic \ energy }{Potential\ energy}=n^2 - 1$

Hence the ratio of the kinetic energy and potential energy possessed by the body = n² - 1.

ANSWER 2:

• Option 1) 9 : 1.

GIVEN:

• A body is projected vertically up with certain velocity.

• It reaches the ground after 12 s

TO FIND:

• The ratio of displacement covered by the body in 2 nd second and 7 th second.

EXPLANATION:

Upward motion:

$\boxed{ \large{ \bold{v = u + at}}}$

v = 0 [ At reaching maximum height it will come to rest ]

It reaches ground in 12 seconds so time taken for half journey = 6 s [ Time of ascent = Time of descent ]

a = - 9.8 m/s² [ As it moves upward ]

0 = u - 9.8(6)

u = 58.8 m/s

$\boxed{ \large{ \bold{S_n = u + \dfrac{a}{2}(2n -1)}}}$

u = 58.8 m/s

a = - 9.8 m/s²

n = 2 [ As we want to find displacement at 2nd second ]

$\sf S_2= 58.8 -\dfrac{9.8}{2}(2(2) -1)$

$\sf S_2 = 58.8 -4.9(4 -1)$

$\sf S_2 = 58.8 -4.9(3)$

$\sf S_2 = 58.8 -14.7$

$\sf S_2= 44.1 \ m/s$

Downward motion:

u = 0 [ At reaching maximum height it will come to rest and it is the starting point for downward motion ]

$\boxed{ \large{ \bold{S_n = u + \dfrac{a}{2}(2n -1)}}}$

u = 0

a = 9.8 m/s²

n = 1 [ As it is the 1st second of downward motion ]

$\sf S_1= 0 +\dfrac{9.8}{2}(2(1) -1)$

$\sf S_2 = 4.9(2 -1)$

$\sf S_2 = 4.9$

$\sf S_2= 4.9 \ m/s$

$\sf \dfrac{S_2}{S_7} = \dfrac{44.1}{4.9}$

$\sf \dfrac{S_2}{S_7} = \dfrac{9}{1}$

Hence the ratio of displacement covered by the body in 2 nd second and 7 th second is 9 : 1.