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# system of two blocks connected with two identical springs and an ideal string is in equilibrium as shown in the figure. extension in each spring in this condition

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### Mohammed

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## The spring block system as shown in figure is in equilibrium. The string connecting blocks A and B is cut. The mass of all the three blocks is m and spring constant of both the spring is k. The amplitude of resulting oscillation of block A isB. 2 mgkC. 3 mgkD. 4 mgk

The spring block system as shown in figure is in equilibrium. The string connecting blocks A and B is cut. The mass of all the three blocks is m and spring constant of both the spring is k. The amplitude of resulting oscillation of block A isB. 2 mgkC. 3 mgkD. 4 mgk Young's Modulus of Elasticity

The spring bl... Question

The spring block system as shown in figure is in equilibrium. The string connecting blocks

A and B

is cut. The mass of all the three blocks is

m

and spring constant of both the spring is

k

. The amplitude of resulting oscillation of block

A is A m g k B 2 m g k C 3 m g k D 4 m g k Open in App Solution

The correct option is B

2 m g k

Just after cutting the string, extension in spring =

3 m g k

The extension in the spring when block is in mean position

= m g k

∴ Amplitude of oscillation

A = 3 m g k − m g k = 2 m g k Suggest Corrections 1 SIMILAR QUESTIONS

Q. The spring block system as shown in figure is in equilibrium. The string connecting blocks

A and B

is cut. The mass of all the three blocks is

m

and spring constant of both the spring is

k

. The amplitude of resulting oscillation of block

A is Q. A block having mass

m and charge q

is connected by a spring of force constant

k

. The block lies on a frictionless horizontal track and a uniform electric field

E

acts on system as shown. The block is relased from rest when the spring is unstretched

( a t x = 0 )

. Identify the wrong statement. Q. The system shown in the figure is in equilibrium. Find the accelerations of the blocks A, B and C just after the string between A and B is cut. All blocks are of equal masses ‘

m

’ each and springs are of equal stiffness. (Assume springs to be ideal and take downward acceleration to be positive).

Q. System shown in figure is in equilibrium. Find the magnitude of net change in the string tension between two masses just after, when one of the springs is cut. Mass of both the blocks is same and equal to m and spring constant of both springs is k. Report n, if answer is mg/n Q. A block of mass

m

is connected to three springs, each of spring constant

k

as shown in figure. The block is pulled by

x in the direction of C

. Find resultant spring constant=? View More

स्रोत : byjus.com

## Two identical springs of constant are connected in series and parallel as shown in figure. A mass m is suspended from them. The ratio of their frequencies of vertical oscillations will be

Click here👆to get an answer to your question ✍️ Two identical springs of constant are connected in series and parallel as shown in figure. A mass m is suspended from them. The ratio of their frequencies of vertical oscillations will be Question

## Two identical springs of constant are connected in series and parallel as shown in figure. A mass m is suspended from them. The ratio of their frequencies of vertical oscillations will be A

B

C

D

## 4:1

Medium Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is C)

We know that, n=

2π 1 ​ m k ​ ​ In series, k= k 1 ​ +k 2 ​ k 1 ​ k 2 ​ ​ = 2 k 2 ​ In parallel, k=k 1 ​ +k 2 ​ =2k ∴,n 1 ​ = 2π 1 ​ 2m k ​ ​ and n 2 ​ = 2π 1 ​ m 2k ​ ​ hence, n 1 ​ :n 2 ​ =1:2 Video Explanation Solve any question of Oscillations with:-

Patterns of problems

>

42 6

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## A mass of 1.5 kg is connected to two identical springs each of force constant 300 Nm^(

A mass of 1.5 kg is connected to two identical springs each of force constant 300 Nm^(-1) as shown in the figure. If the mass is displaced from its equilibri Home > English > Class 12 > Physics > Chapter > Oscillations >

A mass of 1.5 kg is connected ...

A mass of 1.5 kg is connected to two identical springs each of force constant

300N m −1 300Nm-1

as shown in the figure. If the mass is displaced from its equilibrium position by 10cm, then the period of oscillation is Updated On: 27-06-2022 Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Transcript

questions on topic simple harmonic motion a mass of 1.5 kg this is a mass is connected to two identical springs each of force constant 307 in the figure ekadam mass is displaced from its equilibrium position by 10 cm then the period of oscillation is so what happens is that if you have a look at the given configuration this distance is fixed to let us assume they moved by a distance x so what you can see that the extension in this that is Delta X will be equal to the compression in this this this spring so basically what you see that in both of them the amount of extension same to If the amount of extension is same can be replaced is

configuration by this we can replace configuration by this basically what I want to say is that both of the strings are looking like the connected in series but they work in such a way that they are working in parallel so this is this is ke and this is 1.5 kg to the net ke net that is net spring constant of this configuration is equals to ke and King parallel that becomes he plays ke Kuchh To 2K and case 304 become too ke that is 600 Newton per metre so now we found out the unit if you have a look at simple spring block system to have a look it's pretty simple spring block system what is the value of Omega for this kind of

configuration we write Omega in here as under root came I am this is for a simple harmonic oscillator also have a simple harmonic oscillator is the value of k is different so for the given configuration Omega will be Quest to under root ke net Bade M what is the value of k 9600 so what will do will put in the values 600 / mass that is 3 by 2 that is 1.5 kg comes out to be equal to 24 2013 become 20 second inverse now I have to find out the time period of oscillation what is time period of oscillation equals to 2 pi by Omega so equals to 2 into 5 / 2020 b n that is approximately equal to

0.3 14 second so we have found out the time period of oscillation for the given configuration of the strings attached and the option matches that option B is the right answer that is the time period of oscillation for the given question will be equals to 0.314 thank you

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Mohammed 1 month ago

Guys, does anyone know the answer?