A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to a 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is

Step-by-step solution

The frequency of 1   kg block is f = 1 2 π k m = 1   Hz 4 π 2 = k m         ,      m = 1 Spring constant, k = 4 π 2 In parallel connection of springs, the equivalent spring constant is k eq = 2 k . f = 1 2 π   K × 2 8 Frequency of 8   kg block is = 1 2   Hz