Question

The displacement of a S.H.M. doing particle when
K.E. = P.E. (amplitude = 4 cm) is
(a) 2 √2 cm (b) 2 cm
(c ) 1/√2cm
(d) √2 cm

Answer 1

Answer:

opotion A

Explanation:

Kinetic energy,

K.E.=

2

1

mv

2

=

2

1

mω

2

(A

2

−x

2

)

Potential energy

P.E.=

2

1

kx

2

=

2

1

mω

2

x

2

Putting, K.E.=P.E.

2

1

mω

2

(A

2

−x

2

)=

2

1

mω

2

x

2

⇒A

2

=2x

2

x=

2

A

=2

2

cm

Answer 2

Answer:

(a) 2 √2 cm

Explanation:

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