Question

28
Oscillations
21. A trolley of mass 1 kg is connected to two
identical spring of spring constant 100 N/m each.
If trolley is displaced from mean position by 1 cm
then maximum velocity of trolley is
(1) 0.01 m/s
(2) 1.4 m/s
(3) 0.14 m/s
(4) 14 m/s
​

Answer 1

Answer:

the answer is explained above

Answer 2

$v=0.07071\ m.s^{-1}$ is the maximum velocity of the object.

Explanation:

Given:

mass of the object, $m=1\ kg$

spring constant, $k=100\ N.m^{-1}$

amplitude of oscillation, $A=1\ cm=0.01\ m$

Now we have the relation of energy as:

$\frac{1}{2} m.v^2=\frac{1}{2} k_e.A^2$ .........................(1)

we have,

$\frac{1}{k_e} =\frac{1}{k} +\frac{1}{k}$

$\frac{1}{k_e} =\frac{1}{100} +\frac{1}{100}$

$k_e=50\ N.m^{-1}$

Now, using eq. (1):

$\frac{1}{2}\times 1\times v^2=\frac{1}{2}\times 50\times 0.01^2$

$v=0.07071\ m.s^{-1}$

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TOPIC: spring oscillations

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