Question

A body of mass M, executes vertical SHM with periods t1 and t2 when sepa
rately attached to spring A and spring B respectively. The period of SHM, when
the body executes SHM, as shown in the figure is t0 Then



explain ​

Answer 1

2π√m/√k

explanation,

f=ma

f=-kx

from both equations

-kx=ma

-kx=m(-w^2x) (w=angular velocity)

k=mw^2

w^2= k/m

w=√k/√m

t=2π/w

t=2π√m/√k

Answer 2

The period of SHM when the body executes SHM is $t_{1}^{-2}+t_{2}^{-2}$

(D) is correct option

Explanation:

Given that,

Mass of body = M

Time period of spring A = t₁

Time Period of spring B = t₂

We know that,

The time period of spring is

$T=2\pi\sqrt{\dfrac{m}{k}}$

Where, m = mass of body

k = spring constant

Time period of spring A is

$t_{1}=2\pi\sqrt{\dfrac{m}{k_{1}}}$

$t_{1}^2=(2\pi)^2\dfrac{m}{k_{1}}$....(I)

Time period of spring B is

$t_{2}=2\pi\sqrt{\dfrac{m}{k_{2}}}$

$t_{2}^2=(2\pi)^2\dfrac{m}{k_{2}}$...(II)

We know that,

The spring connected in parallel

So, the resultant of spring is

$k_{0}=k_{1}+k_{2}$

We need to calculate the time period of spring

On addition equaton (I) and (II)

$\dfrac{1}{t_{1}^2}+\dfrac{1}{t_{2}^2}=\dfrac{1}{(2\pi)^2}\dfrac{k_{1}}{m}+\dfrac{1}{(2\pi)^2}\dfrac{k_{2}}{m}$

$\dfrac{1}{t_{1}^2}+\dfrac{1}{t_{2}^2}=\dfrac{1}{(2\pi)^2}\dfrac{1}{m}(k_{1}+k_{2})$

$\dfrac{1}{t_{1}^2}+\dfrac{1}{t_{2}^2}=\dfrac{1}{(2\pi)^2}\dfrac{1}{m}(k_{0})$

$t_{1}^{-2}+t_{2}^{-2}=t_{0}^{-2}$

Hence, The period of SHM when the body executes SHM is

Learn more :

Topic : Simple harmonic motion

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