A particle is executing linear SHM. For this match the following :- ( i If mass of the block is doubled (keeping other factor constant) a Energy of oscillation becomes 4 times ii If amplitude of oscillation is doubled

Correct answer: A. (i-c), ii -a, iii -d, iv -b aligned &Energy E=(1)/(2)KA^2=(1)/(2) mw^ 2 A^2 &v_ =A ω=A K m &T=2 π m K & ( P.E. ) _ = (1)/(2)KA^2 aligned ) (i) ( ) Mass is doubled ( ) Time period become ( 2 ) times. (ii) ( ) Amplitude ( ) double Energy ( 4) times (v_ 2) times (P.E.) ( _ 4) times (iii) (K ) double Energy = double (v_ max = 2 ) times (T 1 2 ) times (P.E.) ( _ max =2) times (iv) Angular frequency ((ω= k m )) is halfed means k becomes 1 / 4 times so maximum speed is halfed.

NEET

Exam: NEET
Subject: Physics
Chapter: Oscillations

NEET · Physics · Oscillations

A particle is executing linear SHM. For this match the following :-...

A particle is executing linear SHM. For this match the following :- (

i	If mass of the block is doubled
	(keeping other factor constant)	a	Energy of oscillation becomes 4 times
ii	If amplitude of oscillation is doubled
	(keeping other factor constant)	b	Max. speed of the particle becomes half times
iii	If force constant is doubled
	(keeping other factor constant)	c	Time period becomes \sqrt{2} times
iv	If angular frequency is halved
	(keeping mass constant)	d	Max. potential energy becomes 2 times

)

Options

A(\mathrm{i}-\mathrm{c}), ii -a, iii -d, iv -b
Bi -c , ii -a , iii -b , iv -d
Ci $-a$ , ii $-c$ , iii $-b$ , iv $-d$
Di -c , ii -b , iii -d , iv -a

Correct answer

A. (\mathrm{i}-\mathrm{c}), ii -a, iii -d, iv -b

Step-by-step solution

\begin{aligned} &\text{Energy }\mathrm{E}=\frac{1}{2}\mathrm{KA}^2=\frac{1}{2}

\mathrm{mw}^

\mathrm{ A}^2 \\ &\mathrm{v}_{\max }

=\mathrm{A} \omega=\mathrm{A}\sqrt{\frac{\mathrm{K}}{\mathrm{m}}} \\ &\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{ m}}{\mathrm{ K}}} \\ &

(

\text{ P.E. })

_{\max }=

\frac{1}{2}\mathrm{KA}^2

\end

{aligned}) (i) (\quad) Mass is doubled (\Rightarrow) Time period become (\sqrt{2}) times. (ii) (\quad) Amplitude (\Rightarrow) double Energy (\Rightarrow 4) times (\mathrm{v}_{\max } \Rightarrow 2) times (P.E.) ({ }_{\max } \Rightarrow 4) times (iii) (\mathrm{K} \Rightarrow) double Energy = double (\mathrm{v}_{\text{max }}=\sqrt{2}) times (\mathrm{T} \Rightarrow \frac{1}{\sqrt{2}}) times (P.E.) ({ }_{\text{max }}=2) times (iv) Angular frequency (\left(\omega=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\right)) is halfed means k becomes

1 / 4

times so maximum speed is halfed.

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