A mass m is suspended from a spring of length l and force constant K. The frequency of vibration of the mass is $${{f}_{1}}$$. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is $${{f}_{2}}$$. Which of the following relations between the frequencies is correct [NCERT 1983; CPMT 1986; MP PMT 1991; DCE 2002]

Question

TuteeHUB · Accepted Answer

NA

TuteeHUB · Answer

$${{f}_{1}}=\sqrt{2}{{f}_{2}}$$

TuteeHUB · Answer

$${{f}_{1}}={{f}_{2}}$$

TuteeHUB · Answer

\[{{b}^{2}}

TuteeHUB · Answer

$${{f}_{2}}=\sqrt{2}{{f}_{1}}$$

1.	A mass m is suspended from a spring of length l and force constant K. The frequency of vibration of the mass is \[{{f}_{1}}\]. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is \[{{f}_{2}}\]. Which of the following relations between the frequencies is correct [NCERT 1983; CPMT 1986; MP PMT 1991; DCE 2002]
A.	\[{{f}_{1}}=\sqrt{2}{{f}_{2}}\]
B.	\[{{f}_{1}}={{f}_{2}}\]
C.	\[{{b}^{2}}<4ac\]
D.	\[{{f}_{2}}=\sqrt{2}{{f}_{1}}\]
Answer» E.

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