Question

کرا
a
A block of mass 200 g falls from a height
of 9.8 cm of the pan of a spring balance. The
mass of the pan and spring is negligible. The
block
get struck to the pan and starts
oscillating simple harmonically in the vertical
direction find the amplitude and energy
of
oscillation, force constant of the spring of
Spring balance is 1960 N/m.​

Answer 1

Explanation:

mg(h+y)=

2

1

ky

2

⇒y=

k

mg

±

k

mg

1+

mg

2kh

At equilibrium, mg=ky

0

y

0

=

k

mg

⇒Amplitude A=y−y

0

=

k

mg

1+

mg

2kh

Energy of oscillation is

E=

2

1

kA

2

=mgh+(

2k

(mg)

2

)